1st.  I 
2d. 


John  Swett 


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Prof.  Pattebson  bas  spent  aooiit  »  years  m  prcpamig  tuio  uww.^.  .v  .:8  an 
effort  to  make  the  study  of  Grammar  attractive,  and  to  embody  that  which  is 
really  good  in  the  Language  Lesson  System  with  the  older  and  more  rigid 
rules  of  Grammar.  We  believe  that  the  effort  has  been  successful,  and  that 
this  is  the  best  teaching  hook  on  this  subject  ever  published. 


POINTS  OF  SUPERIORITY 

OF 

SHELDON    &    CO.'S 

MODERN   SCHOOL  READERS 


IN    FIVE    BOOKS,    ELEGANTLY    iLLUSTRATED. 

1st.  They  contain  about  one  third  more  reading  matter,  than 
any  other  series. 

2d.  They  are  bound  in  the  strongest  and  most  durable 
manner.  The  first  three  in  full  cloth  binding;  the  sheets 
being  held  together  by  metallic  plates  (process  patented).  It  is 
impossible,  therefore,  to  have  loose  sheets.  The  Fourth  and 
Fifth  have  cloth  sides  and  leather  backs. 

3d.  Great  pains  has  been  taken  to  give  the  greatest  possible 
variety  of  reading  matter. 

4th.  The  books  embody  the  methods  used  by  the  most  suc- 
cessful teachers  of  reading. 

5th.  The  Phonic  Elements  of  which  familiar  words  are 
composed,  are  presented  one  element  at  a  time. 

6th.    The  script  letters  are  introduced  one  by  one. 

7th.  New  words  are  printed  in  black-faced  type  at  the 
head  of  each  lesson,  in  the  First  and  Second  Readers. 

8ih.  "Memory  Gems"  are  introduced  into  the  Second  and 
succeeding  books ;  Elliptical  Exercises  in  the  Fourth. 

9th.  A  vocabulary  of  all  difficult  words  used  is  placed  at  the 
end  of  the  Third,  Fourth,  and  Fifth  Readers. 

TOth.  They  are  free  from  extremes  of  all  kinds,  and  present  the 
best  methods  of  instruction  and  choicest  selections. 

nth.  In  elegance  of  manufacture  they  cannot  be  surpassed. 
The  illustrations  are  more  elegant  than  those  in  any  other  series. 

I2th.  Their  price  is  very  low,  considering  their  size,  and  the 
elegant  and  durable  style  in  which  they  are  made. 

SHELDON     &     COMPANY, 

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OLNEY'S     NEW     SERIES 


THE 


COMPLETE  ALGEBRA. 


FOR  HIGH  SCHOOLS,  PREPARATORY  SCHOOLS,  AND 
ACADEMIES. 


BT 


EDVV^ARD    OLNEY, 

Professor  of  Mathematics  in  the  University  of  Michigan, 


NEW     EDITIOIT 


NEW    YORK 

SHELDON      &     COMPANY, 

8    MURRAY    STREET. 


0 


OLNEY'S     NEW     SERIES 

EMBRACES  THE  FOLLOWING  BOOKS. 


FIRST  LESSONS  IN  ARITHMETIC,  |  „         „      ,     „      . 
PRACTICAL  ARITHMETIC.  [  Tvv,.-Book  Series. 


This  Series  has  more  examples,  and  at  less  price,  than  any  ever 
published. 


SCIENCE  OF  ARITHMETIC. 
(A  strictly  High  School  Text-Book^ 
Send  for  full  Circular  of  Olney's  Arithmetics. 


FIRST  PRINCIPLES  OF  ALGEBRA. 

COMPLETE  ALGEBRA.    (Newly  electrotyped  in  large  type.) 
NEW  ELEMENTARY  GEOMETRY. 


Copyright,  1870,  1875,  1878,  1881,  by  Sheldon  <&*  Co, 


Smith  &  McDougal,  Ei.ectrotypbbs, 
82  BeekmuD  iSt.,  N.  V. 


^^^ 


^i^^^l^     'i'^  J  ^l^    t    ^i^ 


This  book  is  a  new  edition  of  the  author's  Complete 
Algebra.  The  changes  made  in  the  original  work  are  not 
sufficient  to  preclude  the  use  of  the  old  edition  in  classes 
with  the  new.  The  matter  of  the  new  edition  runs  article 
for  article,  and,  with  very  inconsiderable  differences,  page 
for  page,  with  the  old.  The  changes  made  consist  in  the 
breaking  up  of  the  longer  definitions  and  rules  into  short 
paragraphs,  for  the  convenience  of  teacher  and  pupil  in  the 
class-room,  the  simplification  of  a  few  of  the  more  abstruse 
demonstrations,  the  omission  of  quite  a  large  number  of 
**  suggestions  "  to  teacher  and  to  pupil,  and  the  substitution 
of  simpler  problems  for  a  few  that  teachers  generally  have 
thought  too  difficult.  These  changes,  together  with  the 
larger  and  more  open  page,  and  the  more  beautiful  typog- 
raphy, it  is  thought  will  render  the  book  still  more  accept- 
able to  teachers  than  it  has  hitherto  been. 

In  effecting  this  revision  the  author  has  had  the  aid  of 
several  accomplished  teachers  who  have  been  familiar  with 
the  practical  workings  of  the  book  in  the  class-room.  Of 
these.  Prof.  C.  H.  Churchill,  of  Oberlin  College,  has  read 
the  entire  book  and  offered  suggestions  freely.  With  refer- 
ence to  the  changes  made  in  the  problems,  the  author  has 

543474 


IV  PBEFACE. 

been  guided  by  the  opinion  of  a  large  number  of  practical 
teachers  who  have  used  the  book  for  years. 

The  sanie  thoroughness  in  the  discussion  of  principles, 
comprehensiveness,  philosophic  accuracy,  and  clearness  of 
statement,  and  the  same  careful  adaptation  to  training  the 
pupil  to  think  clearly  and  express  his  thoughts  with  pro- 
priety, characterize  this  edition  as  won  for  the  first  its  great 
popularity.  This  book  is,  therefore,  not  a  mere  child's 
book,  but  is  designed  for  pupils  who  have  the  knowledge 
of  Arithmetic  usually  considered  requisite  for  a  commence- 
ment of  this  study.  For  less  mature  pupils  the  author  has 
prepared  another  volume. 

This  treatise  is  designed  to  lay  the  foundation  of  a  good 
mathematical  education.  Algebra  develops  the  mathemat- 
ical language,  and  is  the  great  mathematical  instrument. 
The  Literal  Arithmetic  (see  pages  1-210)  is  the  basis  of  all 
mathematical  training.  Without  a  good  measure  of  ability 
to  handle  the  various  forms  of  literal  expressions — the  more 
complex  as  well  as  the  more  simple — it  is  impossible  to 
become  a  mathematician.  Hence  an  unusual  amount  of 
care  is  bestowed  upon  this  part  of  the  work.  A  mere 
elementary  knowledge  of  the  simpler  forms  of  numerical 
equations,  is  indeed  easily  obtained,  but  it  is  of  very  little 
worth.  A  thorough  mastery  of  the  elements  of  Algebra 
as  presented  in  this  volume  is  one  of  the  most  valuable 
acquisitions  which  can  be  secured  in  the  schools,  whether 
we  consider  its  value  as  a  mental  discipline,  or  as  a  founda- 
tion for  more  advanced  work. 

The  attempt  to  train  the  pupil  to  methods  of  reasoning, 
rather  than  in  mere  methods  of  operating,  has  given  char- 


PREFACE.  T 

acter  to  the  presentation  of  every  topic.  Propositions  are 
clearly  stated  at  the  outset,  and  demonstrations  are  given 
in  form,  and  with  the  rigor  of  a  geometrical  argument. 
That  there  is  some  defect  in  our  methods  of  instruction,  in 
this  regard,  must  be  painfully  evident  to  every  one  who  has 
been  called  to  examine  large  numbers  of  our  youth  in  this 
study.  The  author  has  examined  for  admission  to  college, 
from  25  to  150  different  students  from  all  parts  of  our  coun- 
try, each  year,  for  the  last  27  years,  and  he  has  almost 
invariably  found  little  or  no  knowledge  of  the  processes  as 
arguments,  even  when  a  good  degree  of  skill  in  the  use  of 
the  processes  had  been  attained.  Perhaps  a  majority  of 
those  examined  could  multiply  the  square  root  of  2  by  the 
cube  root  of  3,  but  scarce  one  in  50  could  develop  the  pro- 
cess in  a  logical  form,  or,  in  most  cases,  give  any  rational 
account  of  it.  Now,  it  need  not  be  said  that,  in  a  course 
of  education,  this  is  a  fundamental  defect;  it  is  failure  just 
where  success  is  vital.  The  processes  of  a  mathematical 
science  are  of  comparatively  little  worth  to  a  great  majority 
of  those  who  study  them  ;  the  development  of  the  reasoning 
powers  to  which  such  studies  are  addressed,  is  of  the  high- 
est importance  to  all.  By  teachers  who  cannot  appreciate 
these  truths,  this  book  will  very  probably  be  misunderstood  ; 
but  to  such  as  do  feel  the  force  of  them,  the  author  appeals 
with  the  fullest  confidence,  not  indeed  that  his  book  will 
meet  the  exigency,  but  that  it  will  be  welcomed  as  an  effort 
in  the  right  direction,  and  as  a  help  in  remedying  this 
radical  defect. 

The  Introductory  portion  is  found  by  many  an  adequate 
discipline  for  the  younger  and  more  immature  pupils  before 


VI  PREFACE. 

entering  upon  the  treatise  proper,  while  the  Appendix 
makes  the  volume  sufficiently  comprehensive  for  many  of 
our  colleges. 

Grateful  for  the  favor  with  which  the  original  work,  as 
well  as  all  he  has  written,  has  been  received  ;  the  author  has 
been  stimulated  to  perfect,  as  far  as  possible,  the  different 
members  of  the  Series,  with  the  hope  of  rendering  them 
still  more  useful. 

EDWARD  OLNEY. 

University  op  Michigan, 
Ann  Arbor,  July,  1881. 


AN    INTRODUCTION    TO   ALGEBRA   ....    1-35 
INTRODUCTION    TO   THIS   BOOK 1-7 

Section  l 

A  BRIEF  SURVEY  OF  THE  OBJECTS  OF  PURE 
MATHEMATICS  AND  OF  THE  SEVERAL 
BRANCHES 1-4 

Section  n. 

LOGICO-MATHEMATICAL    TERMS 5-7 

*(0> 

PART   I. 

LITERAL     ARITHMETIC. 

Chaptei^  I. 

FUNDAMENTAL     RULES. 

Section  i. 

NOTATION 8-27 

Section  ii. 

ADDITION 38-40 


vitt  CONTENTS. 


ECTION  in. 

SUBTRACTION        40-49 

Section  iv. 

MULTIPLICATION 50-62 

SECTfON    V. 
DIVISION 63-76 


FACTORING. 

Section  i. 

FUNDAMENTAL    PROPOSITIONS 77-88 

Section  n. 

GREATEST  OR    HIGHEST  COMMON    DIVISOR   .  88-98 

Section  in. 

LOWEST  OR   LEAST  COMMON   MULTIPLE    .      98-101 

Chaptei|  iil 

FRACTIONS. 

Section  i. 

DEFINITIONS     AND      FUNDAMENTAL     PRINCI- 
PLES         102-106 


CONTENTS.  IX 

ECTJON  II. 

REDUCTIONS 107-117 

Section  in. 

ADDITION       117-121 

Section  iv. 

SUBTRACTION        121-124 

Section  v. 

MULTIPLICATION 124-129 

Section  vi. 

DIVISION 13C-127 

POWERS     AND     ROOTS. 

Section  i. 

INVOLUTION        138-153 

Section  ii. 

EVOLUTION 154-177 

Section  hi. 

CALCULUS    OF    RADICALS 178-192 

Section  iv. 

COMBINATIONS    OF    RADICALS 192-210 


CONTENTS 


PART    II. 

ALGEBRA 


SIMPLE     EQUATIONS. 

Section  i. 

EQUATIONS  WITH  ONE  UNKNOWN  QUANTITY  211-252 

Section  n. 

SIMPLE,  SIMULTANEOUS,  INDEPENDENT  EQUA- 

.      TIONS  WITH  TWO  UNKNOWN  QUANTITIES  253-271 


ECTfON  in. 

SIMPLE,  SIMULTANEOUS.  INDEPENDENT  EQUA- 
TIONS WITH  MORE  THAN  TWO  UNKNOWN 
QUANTITIES 272-280 

ChAPT1I|  II. 

RATIO,  PROPORTION,  AND  PROGRESSION. 

Section  i. 

RATIO     .     . 281-285 

Section  h. 

PROPORTION  285-293 

Section  m. 

PROGRESSIONS 294-313 


CONTENTS.  XI 


BUSINESS      RULES      [OF     ARITHMETIC] 


ECTION    I. 
PERCENTAGE 314-31G 

Section  n. 

SIMPLE  INTEREST  AND  COMMON  DISCOUNT  317-326 


ECTION  III. 

PARTNERSHIP 327-330 


ECTION     IV. 
ALLIGATION 330-334 

<^HAPTEI\  lY. 

QUADRATIC     EQUATIONS. 


ECTION  I. 

PURE    QUADRATICS 335-344 

Section  ii. 

AFFECTED  QUADRATICS       344-356 

Section  hi. 

EQUATIONS  OF  OTHER  DEGREES  WHICH    MAY 

BE    SOLVED   AS   QUADRATICS    ....    357-366 


xii  CONTENTS. 

Section  iv. 

SIMULTANEOUS  EQUATIONS  OF  THE  SECOND 
DEGREE  BETWEEN  TWO  UNKNOWN  QUAN- 
TITIES           367-380 

€haptei|  ¥. 

LOGARITHMS 381-390 

<K»'-*- ■ 

APPENDIX. 
Sectjon  I. 

DIFFERENTIATION. 

DEFINITIONS 391-394 

RULES  AND    EXAMPLES 394-401 

INDETERMINATE    COEFFICIENTS 401 


ECTfON  H. 

THE   BINOMIAL   FORMULA    DEMONSTRATED        .      .      .     402-404 

ECTfON  III. 

THE   LOGARITHMIC   SERIES    PRODUCED        ....    404-409 
PRODUCTION   OF  A  TABLE  OF   LOGARITHMS    .      .      .    409-411 


ECTION    IV. 

HIGHER    EQUATIONS 411-422 

Section  v. 

INTERPRETATION    OR    DISCUSSION    OF   EQUATIONS     .     423-436 

Section  vi. 

PERMUTATIONS  AND   COMBINATIONS      .      .     .  -  .      .    436-439 


Introduction  (^^ 


HOW  LETTERS  ARE  USED  TO  REPRESENT  NUMBERS. 

Ex.  1.  How  many  fives  are  3  times  5  -f  4  times  5  +  2 
times  5 V  Or  3  fives  +  4  fives  +  2  fives ?  Or3x5-f4x5 
+  2x5?  Ans.y  9  times  5,  or  9  fives,  or  9x5. 

2.  How  many  dozen  are  2  dozen  +  8  dozen  +  5  dozen  + 1 
dozen  ? 

3.  How  many  score  are  3  score,  5  score,  8  score,  and 
2  score  ? 

4.  Without  performing  any  multiplications  tell  the  sum 
of  2  times  8,  5  times  8,  and  4  times  8. 

5.  Needles  are  put  up  in  papers,  with  the  same  number 
in  each  paper;  how  many  are  3  papers +  5  papers  +  4  papers 
+ 1  paper  ? 

Suggestion.  — In  each  of  the  above  exercises  a  certain  number  of 
units  is  taken  in  a  group,  the  number  being  the  same  in  eacli  group 
in  any  one  example ;  thus  in  Ex.  1,  the  groups  are  5'8 ;  in  2,  a 
dozen ;  in  3,  a  score ;  in  4,  they  are  8's ;  in  5,  some  certain  numl)€)\ 
hut  we  do  not  need  to  know  what  number. 

NOTE.—TJiis  Introduction  is  designed  for  pupils  ivho  have  not 
studied  the  author^s  **  First  l*rinripleti^^*  and  for  rchom  thr  body  of 
this  work  may  he  deemed  too  severe,  Sonietchat  mature  ^>M/>tfj«,  with 
a  f/ootl  knowledf/e  of  Arithmetic,  will  be  able  to  commence  at  page  1  of 
the  body  of  the  work. 


2  INTRODUCTIOl^. 

i/  irt  t^gebreo  we  often  use  letters  to  represent,  or 
stand  for,  numbers.  The  following  exercises  will  show 
how: 

6.  Two  times  a  certain  niimber -{-'^  ivaiQ^  ihQ  same  nii7n- 
ber  4-  4  times  the  same  number,  are  how  many ;  i.  e.,  how 
many  times  that  number  ? 

7.  Let  a  stand  for  a  certain  number.  How  many  are 
4  times  a-fS  times  «  +  2  times  a  ?  Ans.,  9  times  a. 

8.  If  we  let  771  represent  the  number  of  needles  in  1  paper 
of  needles,  how  many  needles  are  there  in  5  papers  -j-  3  papers 
+  1  paper,  or  how  many  times  m  are  5m's,  3m's,  and  Im? 
If  m  represents  25,  how  many  are  ^m  ? 

2,  Thus  we  may  use  any  letter  to  represent  any 
number,  ojily  so  that  it  always  means  the  same 
number  in  the  same  exercise. 

3,  When  a  letter  is  used  to  represent  a  number,  the 
figure  which  tells  how  many  times  the  jvumber  repre- 
sented by  the  letter  is  taken,  is  just  written  before  the 
letter,  the  woi^d  ''times  "  being  left  out. 

Illustration.  Ba  means  3  times  a,  45  means  4  times  5,  7m  means 
7  times  m,  105a;  means  105  times  the  number  represented  by  a*,  what- 
ever that  number  may  be. 

4,  A  Coefficient  is  a  number  placed  before  a.  letter 
to  tell  how  vvany  times  the  letter  is  taken. 

A  coefficient  is  therefore  a  factor.  Thus  in  5«,  5  and  a  are  two 
factors  of  the  number  represented  by  5a.  The  letters  are  called 
Literal  factors.  The  figures  are  called  the  Numerical  factors.  Thus 
in  5a,  5  is  the  numerical  factor.  Letters  may  be  used  as  coefficients 
as  well  as  figures.  If  no  figure  stand  before  a  letter,  the  letter  is 
taken  once,  or  its  coefficient  is  said  to  be  1.  Thus,  m  means  one 
time  w,  and  its  coefficient  is  1. 

9.  Mention  the  numerical  coefficients  and  the  literal  fac- 
tors in  the  following  expressions : 


HOW    LETTERS   REPRESENT   NUMBERS.  3 

(1.)  6a-\-5b  +  3fu  —  4fi.  (3.)  a—b—VZc. 

(2.)  3«  — 15m  +  22m— 4^  +  ^— ^.      (4.)  d4-2c— m-|-4w— «. 

Queries. — Of  what  is  12  the  coefl5cient  in  12aj?  Of  what  is  x 
the  literal  factor?     Of  what  is  13  the  numerical  factor? 

10.  What  is  the  sum  of  10a-{-ba-[-7a-\-2a^*  If  a  stand 
for  13  what  is  the  sum  ? 

A?is.,  10a -\- 6a -\- 7  a -[-2a  =  24«  =  24x13  =  312. 

Query. — If  the  a  in  10a  meant  one  number,  in  the  5a  another 
number,  the  a's  in  la  and  2a  still  other  nuinlK-rs,  could  you  answer 
this  exercise  in  the  sanu:  way?  You  could  not  answer  it  at  all. 
The  a  must  mean  the  same  number  all  the  time,  in  the  same 
example. 

11.  3a  +  2a4-5rt  +  8rt,  are  how  many  times  «?  Arts.,  18^. 
How  much  is  this  if  a  is  6  ?  Ans*,  108.  How  much  if  a 
is  11?     Ans.,1^%. 

12.  Eleven  times  8,  minus  5  times  8,  are  how  many- 
times  8? 

13.  11a— 5a  are  how  many  times  a?  Eleven  times  any 
number,  minus  5  times  the  same  number,  are  how  many 
times  that  number  ? 

14.  \2x—lx  =  how  many  times  .r?  How  much  is  this 
if  X  represents  3?  II  x  represents  2\?  Ans.  to  the 
last,  11. 

15.  6b  -\-  4b  -{■  10b  —  12b  =  how  many  times  b  ?  How 
much  is  6b-\-4.b  +  10b—12b,  if  b  =  3f  ? 

16.  How  much  is  3?/i  +  8m— 47W  +  67?i— 5?w— 2m?  How 
much  is  this  if  /w  =  -f  ? 

Solution.  Sm  +  Sfn  are  11m,  11m— 4m  are  7m,  7m4-6m  are  1dm, 
13w_57w  are  8;w,  Sm—2m  are  6m.  Hence  the  answer  is  6m.  If 
7«  =  I,  6m  =  6  X  I  =  4. 


♦  The  pupil  is  presumed  to  be  acqnaioted  with  the  use  of  the  signs.    They  are 
explained  in  the  body  of  the  full  treat i(>c',  dimply  as  a  part  of  the  science. 


4  INTRODUCTION. 

5,  In  such  examples  as  the  last  you  can  add  together  all 
the  quantities  with  the  +  sign  into  one  sum,  making  in 
this  case  17w,  and  all  those  with  the  —  sign  into  another, 
making  in  this  case  11m,  and  then  subtract.  Thus,  17m 
—11m  is  6m,  the  same  result  as  before.  This  is,  generally, 
the  better  way  to  solve  such  examples. 

17.  How  much  is  10a:— 4a;— 2:r-f  3a;— 8a;-|- 11a;?  How 
much  is  it  if  a;  is  3  ? 

18.  How  much  is  10a:— ISa*  ? 

Suggestions. — Of  course  we  cannot  take  15a;  out  of  10a:.  But 
we  can  take  lOaj  of  the  15a;  from  the  first  lOu,  and  there  will  then 
remain  5x  of  the  15a;,  which  cannot  be  taken  out  of  the  10a;.  We 
indicate  this  by  writing  it  —5x.  This  means  that  oa*  was  to  be 
subtracted,  but  that  we  had  nothing  to  take  it  from. 

19.  How  much  is  3a; — 6x — 2a:?  Ans.,  — 4a:. 


Query. — What  does  this  answer  mean? 

20.  How  much  is  12a-{-3a—5a—20a?  How  much  is 
this  if  a  =  ^. 

Query. — What  does  the  answer  mean? 

21.  How  much  is  2  times  3  times  a  certain  number,  as  5  ? 

A?is.,  6  times  the  number. 

22.  How  much  is  5  times  7m  ?  that  is,  5  times  7  times  a 
number  which  we  will  represent  by  m  ? 

Ans.,  35  times  m,  or  35m. 

23.  How  much  is  6  times  Sa?  7  times  3a  ?  10  times 
lb  ?  9  times  8y  ?  9  times  8  times  a  number  are  how  many 
times  that  number  ? 

24.  How  much  is  3  times  7m,  and  4  times  8m,  minus  2 
times  22m,  if  m  represents  6  ?  Ans.,  9m,  or  54. 


HOW  LETTEE8   REPRESEin:  NUMBERS.  5 

25.  What  is    10a  divided   by  2 ;   that  is,  what  is  J  of 
10a  ?    272;  divided  by  9  is  how  much  ? 

Ans.  to  the  last,  3x, 

26.  How  many  times  a  number  is  10  times  that  number, 
divided  by  2  ;  that  is,  i  of  10  times  a  number  ? 

27.  How  much  is  J  of  48a:  ?    25a;  divided  by  5  ?     ^  of 
11a;?    11a;  divided  by  11  ?    7x  divided  by  7  ? 

28.  Divide  10a;  by  5,  then  add  3a;,  then  multiply  by  2, 
then  subtract  4a;,  then  divide  by  3  ?    What  is  the  result  ?  . 


M€^M>M 


COMBINATIONS    OF     LETTERS. 

6*.  When  tiuo  lettei's  representing  numbers  are  wriir- 
ten  side  by  side,  their  product  is  indicated. 

Thus,  ab  means  the  product  of  the  two  numbers  repre- 
sented by  a  and  b,  or  a  x  b,  Sabc  means  3  times  the  product 
of  the  numbers  represented  by  «,  b,  and  c,  ov  Sxaxbxc. 

[Note.  Instead  of  saying,  as  above,  the  number  represented  by  a, 
we  usually  simply  say  "  the  number  a,"  or,  "  «,"  without  using  the 
word  number  at  all.  Thus  we  say  3  times  the  product  of  a,  6, 
and  c] 

7,  If  we  want  to  represent  the  product  of  a  number 
represented  by  a  letter,  as  a,  by  itself  a  certain  numJber 
of  tijnes,  instead  of  uTiting  aa,  or  aaa,  etc.,  as  we 
might,  we  write  €^,  a^,  etc. 

Thus  ¥  means  the  same  as  bbbb.  d^  is  read  "a  square ; '' 
a\  " a  cube ;"  ¥,  "  b  fourth  power ;"  a^,  "  x  fifth  power ;"  etc. 

[Do  not  read  &*  "h  fourth,''^  but  "&  fourth  power."  You  will 
hereafter  learn  that  6'"  is  *'5  fourth,"  and  is  quite  a  different  thing 
from  b*.] 


•  IinDKODUCTIOK. 

The  little  figure  placed  at  the  right  and  a  little 
ahove  the  letter  is  one  form  of  what  is  called  an 
Exponent. 

Thus  in  m"^,  2  is  the  exponent  of  m,  and  m"  means  m  times  m. 
In  6aj',  3  is  the  exponent  of  x^  and  6x^  means  Qxxxxxx. 

[An  exponent  does  not  always  indicate  a  product.  It  does  this 
only  when  it  is  a  whole  number.  Thus  a~^  does  not  mean  aaa ; 
and  in  a%  the  expcment  f  has  quite  a  different  meaning  from  the  3, 
or  —3,  above.  These  other  meanings  of  exponents  will  be  ex- 
plained later  in  the  work.] 

Ex.  1.  How  much  is  4a^h,  ii  a  —  2,  and  b  =  5?  How 
much  is  Sa^ly^x,  if  a  =  3,  b  =  2,  x  =  S?  How  much  is 
a^by,  it  a  =  2,  b  =  1,  tj  =  S? 

2.  How  much  is  aW  +  2ay  —  by,  if  a  =  4,  b  =  3, 
y  =  2  ?  How  much  is  3a'^by^—2ay'^-\-5b,  the  letters  having 
the  same  values  as  before.  How  much  is  5by—2ai^-{-4:a^y^ 
—2a? 

3.  How  many  times  ab^  is  4:ab^  -\- 2ab^  —  3ab^  ?  How 
many  times  a^y  is  10a^y-\-4:a^y—6a^y—a^y  ? 

4.  How  many  times  amy  is  4  times  3amy  ?  How 
much*  is   6   times   2amy?     4  times   'ila^b^c^?     10   times 

5.  How  many  times  ax  is  -J  of  20ax?  |  of  S5ax? 
W2ax  divided  by  3  ?  How  much  is  J  of  72a^a^?  125^5^2 
divided  by  25  ?     18ny^  divided  by  9  ? 

S.  We  have  learned  in  arithmetic  that  representing 
numbers  by  the  figures  1,  2^  3,  4,  etc.,  is  called  Arabic  or 
Decimal  Notation.     In  like  manner,  representing  numbers 

by  the  small  letters  of  the  alphabet,  as  a,  b,  c,  d, x, 

y,  etc.,  is  called  Literal  Notation. 


*  This  means  the  same  as  the  preceding  question. 


COMBINATIONS    OF    NUMBERS.  7 

The  pupil  will  see  that  this  Literal  Notation  is  altogether  a 
diflferent  thing  from  the  Roman  Notation,  in  which  the  seven  capital 
letters,  I,  V,  X,  L,  C,  D,  M,  are  used.  Because  the  Literal  Notfition 
is  so  much  used  in  Algebra,  it  is  often  called  the  Algebraic  Notation. 
But  this  notation  is  just  as  much  used  in  some  other  branches  of 
Mathematics,  as  in  Algebra. 

f>,  A  Term  is  a  group  of  letters  combined  together  by 
multiplication,  or  division,  or  both. 

In  the  expression  5a'a;  —  3a  + r- ,  oa'j;,  3a,  — ,  and   -^, 

are  terms. 

A  Monomial  is  an  expression  consisting  of  but  one 
term. 

A  Binomial  is  an  expression  consisting  of  two  terms 
connected  by  the  sign  +,  or  — . 

A  Trinomial  lias  three  terms. 

A  Polynomial  is  an  expression  with  more  than  one 
term. 

6.  Point  out  the  monomials,  binomials,  trinomials,  and 
polynomials  in  the  following :  2ax—^b^,  6xy—0cd-\-a—2yy 
dahnHy,  (?—d},  a  +  m,  a  +  b  +  c—d,  226a^t^c^d^,  abcd,a^b, 
ab,  c-x^y  +  ax,  x^+y'^,  \W  +  ^xy,  Q?-y\  Sahnxy,a^-5x+d, 

10,  Similar  Terms  are  terms  which  have  the  same 
literal  factors  affected  with  the  same  ei^onents. 

7.  Point  out  the  similar  terms  among  the  following: 
3ax%  2ax,  -5^2^,  ax,  lUx\  16cy\  -12(^y%  Sa^x^,  -5cy% 
Qcy^  lOax^,  c^y^  Sa^x,  dax^,  61a^x. 

11.  Positive  Terms  are  terms  which  have  the  sign  -|-. 
Negative  Terms  are  terms  which  have  the  sign  — . 
When  no  sign  is  expressed,  4-  is  understood. 


nrrRODucTioiir. 


HOV\/    NUMBERS    ARE    ADDED    IN    THE    LITERAL 
NOTATION. 

12.  Rule. — I.  Write  the  expressions  so  that  similar 
terms  shall  stand  in  the  same  column. 

II.  Comhine  the  teriyis  in  each  column  and  tvrite 
the  result  underneath  with  its  own  sign. 

The  polynomial  thus  found  is  the  sum. 

Ex.  1.  Add  5ax—2cy,  3ax-{-^cy,  cy—2ax,  —^ax—dcy, 
—ax  +  6cy,  and  2ax-{-2cy. 

Operation. — Having  written  the  numbers  so  5ax  —  2cy 

that  similar  terms  fall  in  the  same  column,  we  3^^    1    ^^y 

may  begin  to  add  with  any  column  we  choose.  9^1.         * 

Adding  the  right-hand  column  we  find  it  makes  '             ^ 

+  7cy,  and  write  this  sum  underneath  the  column  ^^^       '^^^ 
added.    In  like  manner  the  other  column  makes     —    ax  -{-  6cy 

Sax  (or  +dax),  which,  as  it  stands  first,  we  write  2ax  ~\-  2cy 

without  any  sign,  as  4-  will  then  be  understood.  o  ^    ,    w~ 
The  sum  is  Sax + ley. 

(3.) 
10am  —     Sdy^  +     2«% 

—  6am  +  4:dy^  —  lOa^x 
^a?n  —  8dy^  —  2d^x 
lam  —  13dy^  +     Ga^x 

—  9am  -f     2dy^  —     ^urcr 
am  +   18dy^  —       a^x 

^am  —  IWx 

bed  +     2a  +     bxy 

♦  W^exx  no  sign  is  expressed  before  a  term,  +  is  understood. 


(2.) 

bed  — 

2a   + 

^xy 

2cd   + 

3a  — 

J>xy 

8a  — 

2xy 

Qcd 

+ 

Uxy 

— 

3a   — 

7xy 

+ 

11a  4- 

xy 

Acd  — 

15a 

StJBTRACTlON  IN  THE  LITERAL  NOTATIOK.  9 

4.  Add  5x—'6a-\-b-^7  and  —4:a—dx-\-2b—d. 

Sum,  2x—7a-^3b—2. 

5.  Add2a  +  3Z>— 4c— 9  and5a— 3^>H-2c— 10. 

6.  Add  Sa-^2b—5,  a  +  bb—c,  and  6fl— 2c  +  3. 

7.  Add   6a;?/— 12a^,    —4:X^  +  Sxy,    Lx^—'Zxy,    and    — 3:^?/ 
-f-4a;l  aS'^^w,  ^xy—^y^. 

8.  Add    3«2  +  4^c— e2-|-l0,     _5a3_|-6*c  +  2e2— 15,     and 
_4a2_9^c_10^>2_^21. 

9.  Add    5a;^«/— 3ca;,    ^7?y,    Ibex,    Idcy—ab,^  llab—12cx 
—^x^y,  and  — 10«J— 13c^. 

10.  Add  'lax-^-^cy — bax—cy-\-3cy—ax-\-llax—by-\-3c. 

11.  Add  —3coi^-\-bay,  ax—Sy,  and  b—dx^ 


ECTION  IVo 


HOW  NUMBERS  ARE  SUBTRACTED  IN  THE    LITERAL 
NOTATION. 

IS,  Rule. — Change  the  signs  of  the  terms  in  the 
subtrahend,  or  conceive  them  to  he  changed,  and  add 
the  result  to  the  minuend. 

Ex.  1.  From  bby—^a^-\-Zofi  subtract  2by-\-Za^+3^. 
Operation. 

bby  —  W-\-30l^     Minuend. 

—  2by  —  Zcfi  —    X^     Subtrahend  with  signs  changed. 

17         fv   2  I   o^  i  "^^  Remainder  pought,  which  is  the  snm  of  the  mintw 
oOy —  vCl^  -f-  /C.'^  \      end  and  the  subtrahend  with  its  signs  changed. 

2.  From  3ax  +  bl/^y^—2m^  take  Sax—bY—^f^^- 

Bern.,  6^/+wi«. 


10  INTBODUCnON. 

3.  From  10a—'db-j-2c—a^  subtract  J--5c  +  a^. 

Rem.,  10a-4:b-\'7c'-2a^. 

4.  From  12xy—3c^  +  ab  take  Qxy-\-c'^—2ab. 

6.  From  bc^y—3ab  take  mx—2c^y. 

Rem.,  Hc^y—Sab—mx, 

6.  From  x^— 11  xyz-i- 3a  take  —6xyz-\-'7—2a—bxyz. 

Rem.,  x^-\-5a—7. 

14,  When  there  is  a  term  in  the  subtrahend  which  has 
no  similar  term  in  the  minuend,  we  see  that  this  term 
appears  in  the  remainder  with  its  sign  changed. 

7.  From     6a(^—2by^  +  4:X—3cy  take    —2aci-\-3b^y-dx 
—3cy-k-m.  Rem.,  Sa(P—2by^—3b^y-\-7x—m. 

8.  From  Sm^x^—3a—4:b  take  a^—b^. 

Re7n.,  8m8ir3— 3«— 4Z>— a2-i-52. 

9.  From  a^-\-2ab  +  b^  take  a^—2ab-\'W. 

10.  From  aH  ^  take  a^—1^. 

11.  From  3cm — y  take  2b— 3c, 

12.  From  x^^2xy^l  take  ^y. 


HOW    NUMBERS   ARE    MULTIPLIED  IN  THE   LITERAL 
NOTATION. 

15.  To  multiply  two  Monomials  together. 

Rule. — /.  Multiply  the  numerical  coefficients.  To 
this  pj^oduct  affix  the  literal  factors,  affecting  each 
letter  iiith  an  exponent  equal  to  the  sum  of  the 
exponents  of  that  letter  in  both  terms. 


MULTIPLICATION   IN  THE  LITERAL  NOTATilON.  11 

//.  If  the  signs  of  the  terms  are  alike,  the  sign 
of  the  product  is  +  ;  if  unlike,  — . 

Ex.  1.  Multiply  6aa^  by  da^x\ 

Operation. — The  product  of  the  numerical  coefficients 
8  and  5  is  15.     To  thi'  we  affix  the  letters  a  and  x;  and        5^,1 
as  a  has  an  exponent  1*  in  the  multiplicand,  and  2  in  the        ^n2r< 
multiplier,  we  give  it  an  exponent  3  in  the  product,  and 


X  an  exponent  5  for  a  like  reason.     The  product  is  there-      ^^^  -^ 
fore  15  aV. 

2.  Multiply  lOmn^  by  S?nhi^.  dxy  by  ix^if.  Hex  by  Sex. 
2a  by  a^. 

3.  Multiply  —  5fl«  by  6b.  Product,  —30a^. 

4.  Multiply  —  3«2a;  by  —2ah?y.  Product,  (ja^x^. 

5.  Find  the  products  of  the  following :  —la^x  by  2axi/; 
lOc^nix^  by  —dm^x;  9a  by  4^» ;  —am  by  —xy;  Sc'^d^  by 
—ab;  —5xy  by  —x^y^;  dax^hy  —5a^;  —Hxy  by  —10a^y\ 
5abx  by  —ca^;  ax  by  cy. 


16,  To  multiply  two  factors  together  when  one  op  both 
are  Polynomials. 

Rule. — Multiply  all  the  terms  of  the  TnultipUcand 
hy  each  term  of  the  multiplier,  and  add  the  pro- 
ducts. 

Ex.  1.  Multiply  2a^x—Uy-\-^m  by  ^a^m. 

Operation.  —  It   is   immaterial  2a^x — ^by-\-^m 

where    we    write    the    multiplier,  2a^l^m 

but  we  may  as  well  write  it  as  in  ~              ~          ^  „_„    " 

arithmetic.    So  also  it  makes  no  ia^V^'nx-Qa^lhny +  ^a?l>'w? 

difference  whether  we  begin  at  the 


«  When  no  exponent  is  expressed,  1  is  always  understood. 


1^  *  tKTRODUCTlOK. 

right  hand  or  the  left  to  multiply.  It  is  cu8to7uary  to  arrange 
the  letters  in  a  term  alphabetically  ;  thus  we  write  ^a^'V-mx  instead  of 
^"^a^xm^  or  any  such  foriii.  There  would  be  really  no  difference  in 
the  value  of  the  terms,  however,  in  whatever  order  the  letters 
came. 

2.  Multiply   6ax—5a^x-i-Saa^  by   2a^x^;    2my—3cy^  by 
jm'c^;  4:ab—3cd-{-x  hy  ay ;  a-\-b~c  hj  x\  lOa^m^x- 
\-Smx^  by  6«y, 

3.  Multiply  Sa^x^—2ay^  +  6xy  by  a^—2xy. 

Operation. 

Sa'^x' — 2ay^  4-  5xy 
a''—2xy 


Product  by  a\  3a V — 2aY  +  5a'«y 

Product  by  —2xy,  —  Qa^x^y  +  ^axy* — lOx^y^ 

Sum  of  partial  prod's.,  3a  V — 2aY  +  ^a'xy—Qa^x"y-\-4:axy*  —  lQx'^y^ 

There  being  no  similar  terms  in  these  partial  products,  we  can 
add  them  only  by  connecting  them  with  their  proper  signs. 

4.  Multiply  x^—2xy-{-y^  by  2x—3y, 

Operation. 

x^—2xy+y^ 
2x—3y 


Product  by  2a;,  2x^—4:x'^y  +  2xy^ 

Product  by  —dy,  —  3a-V  +  6ry"— 3y° 

Entire  product,  2x^^7x'y+Sxy^—Sy^ 

5.  Multiply  a-{-n  hy  a—n.  Prod.,  a^—n\ 

6.  Multiply  a'^+a^  +  a^  +  a-r-l  by  a^—l. 

7.  Multiply  a^—2ah  +  J^  by  a^-\-2al)-\-l^. 

8.  Multiply  m-\-7i  by  m-\-n. 

9.  Multiply  m—n  by  m—n. 
10.  Multiply  4:X^—9y^  by  x-{-2y. 


DIVISION  IN  THE  LITERAL  NOTATION.  13 

11.  Multiply  together  x—3,  x—b,  a:— 4,  and  x—1. 

Prod.,  a:4_i9a,^+i3ia45_389a;^42o. 

12.  Multiply  x^-\-y^-\-z^—xy—xz—yz  by  x-\-y-^z. 

Prod.,  7fi-\-^-\-z^—3xtjz. 


E€TION   ¥L 


HOW    NUMBERS    ARE    DIVIDED    IN    THE    LITERAL 
NOTATION. 

17.  To  divide  Monomials  when  dividend  and  divisor 
consist  of  the  same  letter  affected  with  exponents. 

Rule. — Subtract  the  exponent  of  the  divisor  from 
that  of  the  dividend.  The  common  letter  with  this 
difference  as  an  exponent  is  the  quotient.  If  the 
signs  of  the  divisor  and  dividend  are  alike,  the 
sign  of  the  quotient  is  +  /  if  unlike,  —. 

Ex.  1.  Divide  «»  by  a\  QiwL,  aK 

The  student  will  observe  that  the  product  of  the  divisor  and 
quotient  must  always  equal  the  dividend.    In  this  case,  «'  x  a^=a\ 

2.  Divide  —x'^  by  x^.  Quot.,  —a^. 

3.  Divide  — m^  by  —  m^.  Quof.,  m. 

4.  Divide  J«  by  —b*.  Quot.,  —P. 

5.  Give  the  quotients  in  the  following  cases :  y'^-^y* ; 
—x^-i-x^y  cfi-, — flS;   — c^h — c^. 


18.  To  divide  one  Monomial  by  another  when  there  is 
no  letter  in  the  divisor  which  is  not  in  the  dividend,  and 
no  exponent  in  the  divisor  is  greater  than  the  exponeni 
of  the  same  letter  in  the  dividend. 


14:  INTEODUOTION. 

Rule. — Divide  the  numerical  coefficients,  and  to 
the  quotient  annejc  the  literal  factors,  affecting 
each  with  an  exponent  equal  to  its  exponent  in 
the  dividend  minus  that  in  the  divisor,  and  sup- 
pressing all  factors  whose  exponents  are  0.  If 
bhe  dividend  and  divisor  have  liTce  signs,  the  quo- 
tient will  have  the  sign  +  ;  if  unlihe,  — . 

Ex.  1.  Divide  Ibl^x^y  by  ^hxy.  Quot.,  Wx, 

2.  Divide  21a^m^ny^  by  —'la^mny.  Quot.^  —Sm^y. 

3.  Divide  — 105ay  by  —21aK  Quot.,  5ayK 

4.  Divide  —ISmn'^x  by  Cjmnx.  Quot.,  — 3w. 
Give  the  quotients  in  the  following : 

5.  12aY  -^  SayK 

6.  —  64ay  _i_  iQay, 

7.  Slx^y^  -^  -  2Wy\ 

8.  —  ZbaWT?  -. UV^x^. 

9.  24my  -, 8my. 


19.  General  rule  for  dividing  one  Monomial  by  another. 

Rule. —  Write  the  divisor  under  the  dividend  in 
tlie  form  of  a  common  fraction,  and  then  reduce 
this  fraction  to  its  lowest  terms  hy  cancelling  all 
factors  common  to  both  numerator  and  denom- 
inator. 

The  sign  of  the  quotient  is  determined  as  in  the 
last  case. 

10.  Divide  l^a^a?  by  4«W 

Operation.    18aV-7-4<*V  =  -t— ,-v  •    Now  in  18  and  4  there  is 

4a^a:^ 

a,  common  factor  2  which  can  be  cancelled ;  the  a^  in  the  numerator 


DIVISION    IN   THE   LITERAL   NOTATION.  15 

will  cancel  two  factors  of  a  from  a*  in  the  denominator,  and  leave 
a  factor  a  in  the  denominator,  and  in  like  manner  a?'  in  the  denom- 
inator cancels  x^  from  the  x^  in  the  numerator,  leaving  x  therein. 


Hence, 

18aV  _  9ar 
'  4rtV  ~  2a' 

11. 

Divide  12my  by  Idmhj. 

Operation. 

5m ' 

12. 

Divide  ^Sa^3?f  by  —Z^a^u^y'^. 

^'""•'  -2' 

13. 

Divide  ^^()}^a^y  by  — 40Z^a;*. 

Quo.,  l 

14. 

Divide  Qax  by  4J)y. 

<?-.  s 

15. 

Divide  —limy  by  lOwa;. 

^-•'-^^ 

Ex's  16  to  21.  Find  the  quotients  in  the  following  cases: 
24^5^:2^15,^.  _21arir2  by  — 14«???2;  —bahhybxy-j  IQa^y^ 
by  —12aY ;  ^a^ips  by  —dx^ ;  — 4w27j3  by  8mw. 


20.  To  divide  a  Polynomial  by  a  Monomial. 

Rule. — Divide  each  term  of  the  dividend  by  the 
divisor ;  and  write  the  quotients,  connecting  them 
with  their  own  signs. 

22.  Divide  Qa'^o(^f—na^3^f-\-lbah^f  by  Zah?y\ 

Operation. 

^^xY )  6aVy''-12aVy''4-15aVy' 
2xY—iaxy*  +  5a''x^y 

23.  Divide  12flY— 16ay  +  20aY— 28ay  by  —  4ay. 

24.  Divide  lba^C'-20acy^-\-bcd^  by  — 5a&c. 

^       '  h       au 


16  INTRODUCI'IOI?. 

25.  Divide  lQm^x^—24:m^x^-{-10m^x^  by  2m^af^. 

26.  Divide  35a^%2^28«s^y— 49ajy  by  —naV^yK 


21.  To  perform  division  when  both  dividend  and  divisor 
ape  Polynomials. 

Rule. — I.  Arrange  both  dividend  and  divisor  with 
reference  to  some  letter  found  in  both,  i.  e.,  place 
that  term  first  at  the  left  which  has  the  highest 
exponent  of  this  letter,  the  term  containing  the 
next  highest  exponent  of  this  letter  next,  etc. 

II.  Divide  the  first  term  of  the  dividend,  by  the 
first  term  of  the  divisor  for  the  first  tei^m  of  the 
quotient. 

III.  Multiply  the  divisor  by  this  term  of  the 
quotient,  and  subtract  the  product  from  the  divi- 
dend. To  this  remainder  bring  down  as  many 
terms  of  the  dividend  as  may  be  necessary  to  form 
a  new  dividend. 

IV.  Divide  the  first  term  of  this  new  dividend  by 
the  first  term  of  the  divisor,  and  proceed  as  before, 
repeating  the  process  till  the  dividend  is  all  used. 

Ex.  1.  Divide  Qa^x^—^a7^—4.aH-\-7:l^  +  a'^hj  —%ax-^d^-\-x\ 

Operation. — Arranging  dividend  and  divisor  vnth  reference  to 
a.^  and  proceeding  according  to  the  rule,  we  have  the  following : 

— 2a^a;  +  ^a?x^ — ^ax^ 
—  2a'a!  +  4aV  — 2ax» 


a'^x'^—2ax^-\-x^ 
aV  — 2aa;Hir* 

The  polynomials  might  equally  well  have  been  arranged  with 
reference  to  a;.  Thn^  x^—2ax-\-a^)x*—^ax^-\-Qa'^x'^—Wx+a*.  Tnis 
arrangement  would  give  x^—2ax-\-d^  for  the  quotient,  which  ia 
essentially  the  same  as  before. 


A    LITTLE    ABOUT    FACTORING.  17 

2.  Divide  x'^f+x^-\-y^  by  xy^x^-\-y\ 

Qiiot,  x^—xy+f, 

3.  Divide  a^-^'Zah-\-l^  by  a-\-b. 

4.  Divide  a^—2ab-{-P  by  a—b. 

6.  Divide  Aax -{- iiX"^ -\- a^  by  2x-\-a. 

6.  Divide  fl2^  +  ^>5-|  a^t^  +  a^  by  fl2— aJ  +  Jl 

7.  Divide  lOac^-'dd^-i-^a^  +  ^i^  +  Sab  +  Sbc  hj  2b+a  +  'Sc. 

8.  Divide  x^—y^  by  ic— y. 

Operation. 

a5*y-y* 


xY-y' 


«y' 


a;y*— y* 

9.  Divide  x'^—y'^  hj  x-\-y. 

10.  Divide  ^a^—b^  by  2rt— J. 

11.  Divide  20ax^-{-4:a^-^x^—2ba^x^  by  2«3-f-2a;8— 5fl^. 

12.  Show  that  (a^-b^)^(a^^ab  +  i^)  =  a—b. 


E€T50M  YII. 


A    LITTLE    ABOUT    FACTORING. 

22.  The  Factors  of  a  number  are  those  numbers 
which  multiplied  together  produce  it. 

Thus  3  and  4  are  the  factors  of  12,  because  3x4  =  12. 


18  INTRODUCTION. 

So    a-\-b    and    a — b   are  the  factors  of  a^—H^,   because 
(a+d)x{a—b)  =  a^—l\    Try  it,  and  see. 

Ex.  1.  What  is  the  product  of  a  and  l  ?  What  are  the 
factors  of  ab  ? 

2.  What  is  the  product  of  3,  x,  and  y  ?  What  are  the 
factors  of  ^txy  ? 

3.  What  does  a^  mean?  What  are  the  factors  of  «8V 
Whatof  «3^2?     OihaW^ 

Suggestion. — Such  an  expression  as  5a^Z>'^  may  be  resolved  in  a 
great  variety  of  ways :  thus  5,  a,  «,  a^  b,  and  h  are  its  factors ;  also, 
5,  a^,  and  5^ ;  also,  5,  a'\  a,  and  6^ ;  also,  5a,  a'',  &  and  5,  etc. 

4.  What  is  the  product  of  a  and  x-\-y  ?  What  are  the 
factors  of  ax-\-ay? 

6.  What  is  the  product  of  Sa  and  a—b?  What  are  the 
factors  of  Sa^—Sab?    Of  2a—2ab  ? 


THE  SQUARE  OP  THE    SUM  OF  TWO  NUMBERS. 

Ex.  1.  What  is  tjie  product  ot  a-\-b  and  a-\-b?  What 
are  the  factors  of  a^  +  2ab  +  ^  ? 

2.  What  is  the  product  of  x-\-y  and  x-\-y'^  What  are 
the  factors  of  x^  +  2xy  +  y"^  ? 

3.  What  is  the  product  of  1+a;  and  l-\-x't  What  are 
the  factors  of  1  +  2a;  +  ic^  ? 

23»  We  see  from  the  last  examples  that  the  square  of 
the  sum  of  two  numbers  equals  the  square  of  one  of 
them,  -\-  twice  the  product  of  the  two,  -\-  the  square 
of  the  second.  Thus  {a-\-b)x  {(t-\-b)  is  the  square  of  the 
sum  of  the  two  numbers  a  and  b,  and  is  equal  to 
a^-\-2ab  +  lP. 

This  principle,  and  those  in  24  and  25,  are  of  great  im- 
portance in  factoring. 


A    LITTLE    ABOUT    FACTORING.  19 

THE    SQUARE    OF    THE    DIFFERENCE    OF    TWO 
NUMBERS. 

Ex.  1.  What  is  the  product  of  x—y  and  x—y  ?  What 
are  the  factors  of  x^ —"Ixy -\;  fi 

2.  What  is  the  product  of  m^n  and  m—n'i  What  are 
the  factors  of  w^—'lmn-^-rfi  ? 

3.  What  is  the  product  of  \—x  and  \—x't  What  are 
the  factors  of  l-'2x-\-a^? 

4.  What  is  the  product  of  2—x  and  2—x?  What  are 
the  factors  of  4— 4a;  +  a;2  ? 

24.  From  these  examples,  we  see  that  the  square  of 
the  differetice  of  tiuo  iiumhers  is  equal  to  the  square 
of  otie  of  them,  —  tivice  the  product  of  the  two,  +  the 
square  of  the  other.  Thus  {x—y)  x  {x—y)  =x-—'lxy-\-y^. 


THE  PRODUCT  OF  THE  SUM  AND  DIFFERENCE 
OF    TWO    NUMBERS. 

Ex.  1.  What  is  the  product  of  x-^y  and  x—y't  What 
are  the  factors  of  x^—y^'t 

2.  What  is  the  product  of  «4-J  and  a—h'-  What  are 
the  factors  of  a^— ^  ? 

3.  What  is  the  product  of  l  +  a;  and  \—x'>  What  are 
the  factors  of  \—x^'i 

4.  What  is  the  product  of  %-\-x  and  %—x'i  What  are 
the  factors  of  4— rc2V 

2*^,  We  see,  from  these  examples,  that  the  product  of 
thc:  sum  and  difference  of  two  numbers  is  equal  to  the 
difference  of  their  squares.  Thus  {x  +  y)x  {x—y)  =:.7?—y\ 


Ex.  1.  What  are  the  factors  of  2«— 25? 

2.  What  are  the  factors  of  ^a^—^a^x  ? 

3.  What  are  the  factors  of  c^+^crf+f?^? 


)  INTRODUCTION. 

4.  What  are  the  factors  of  o^ — 'lam-{-tr?  ? 

5.  What  are  the  factors  of  c^—c^  ? 

6.  What  are  the  factors  of  a;^— 2a;  +  l  ? 

7.  What  are  the  factors  of  9— ic^? 

8.  What  are  the  factors  of  a2^2a+l  ? 


MCTION   VIIL 


HOW  OPERATIONS   IN  FRACTIONS  ARE   PERFORMED 
IN   THE    LITERAL   NOTATION. 

2(y.  For  the  various  operations  in  fractions  in  the 
literal  notation^  the  ordinary  rules  of  arithmetic  for  the 
corresponding  cases  apply,  only,  that  the  fundamental 
operations  of  addition,  subtraction,  multiplication,  and 
dinsion  are  performed  by  the  preceding  rules. 


TO  REDUCE  FRACTIONS  TO  THEIR  LOWEST 
TERMS. 

^7.   What  is  the  rule  for  this  operation  in  arith- 
metie  ? 

Ex.  1.  Reduce  —^ — ^—^  to  its  lowest  terms.  Result,  ^ 

105W 

2.  Reduce    ^  ^   •  „  to  its  lowest  terms. 

3.  Reduce  -tt^^- ^— „„  to  its  lowest  terms. 

Suggestion. — Divide  numerator  and  denominator  by  4<z'6. 

4.  Reduce ^- — ^^  to  its  lowest  terms. 

Wy 

5.  Reduce vvoq  o— tjt-tz —  to  its  lowest  terms. 


FRACTIONS    IN    THE    LITERAL    NOTATION.  ^1 

6.  Keduce  -5 — -„  to  its  lowest  terms. 

a^  -\-  'lax  +  7?' 

Suggestion. — Try  a\x  and  see  if  it  will  not  divide  both  terms 
of  the  fraction. 

I  1.  -J 

7.  Reduce  -„ — 7.,  to  its  lowest  terms.         Result, 


8.  Reduce  — ^ — -„  to  its  lowest  terms. 

Suggestion. — Will  any  monomial  divide  the  terms  of  the  fraction  ? 


IMPROPER   FRACTIONS    REDUCED    TO   WHOLE 
OR   MIXED    NUMBERS. 

28,   What  is  the  rule  for  this  operation  in  arith- 
metic? 

Ex.  1.  Reduce to  an  integral  form. 

Result,  x—d. 
Suggestion. — Divide  the  numerator  by  the  denominator. 

2.  Reduce r -^ to  an  integral  form. 

%a—dx 

3.  Reduce ' -.-1, to  an  mtesral  form. 

^c^—ax 

4.  Reduce to  an  integral  form. 

oa  +  x 

6.  Reduce ■ ^r— -,^ — ^ to  an  integral  form. 

6.  Reduce  — —  to  an  intesral  or  mixed  form. 

y 


22  INTRODUCTION. 


Operation.    ^L^ —       The  term  ay  can  be  divided  by  y,  giving 
a  +  - 

y 

a.     But  we  can  only  indicate  the  division  of  5  by  y,  by  writing  it  in 
the  form  - ,  and  as  the  sign  of  both  b  and  y  is 4-,  'he  quotient  -  is  +, 

y  y 

and  is  to  be  added  to  a. 

20a;S 10a;  4- 4 

6.  Reduce to  an  integi'al  or  mixed  form. 

oX 

4 

Result,  4:^2—2-1-—-. 
ox 

2w^x  — ~  ic^ 

7.  Reduce to  an  integral  or  mixed  form. 

Result,  2ax 

a 

Query. — "Why  has  —  the  —  sign? 

/rS  _j_  a;3 Q^ 

8.  Reduce to  an  integral  or  mixed  form. 

a  +  x  ^ 

X* 

Result,  a^—axA-x^ 

a-\-x 

Sax  4-  W 

9.  Reduce to  an  integral  or  mixed  form.     Also, 

X 


ab-y^ 


MIXED    NUMBERS    REDUCED    TO    IMPROPER 
FRACTIONS. 

2^.    JVJiat  is  the  rule  for  this  operation  in  arith- 
metic ? 

Ex.  1.  Reduce  a —  to  the  form  of  a  fraction. 
a 

Operation. — The  integral  part  is  a.     This  multiplied  by  a  makes 
a^     From  this  &  is  to  be  subtracted,  because  of  the  —  sign.    Thus 

we  have  for  the  result . 


FRACTIONS    IN    THE    LITEEAL    NOTATION.  23 

2.  Reduce  2x4-—  to   the  form   of  a  fraction.      Also, 

X 

3a 
a—x 

X 

3.  Reduce  a4-x — to  the  form  of  a  fraction. 

a—x 

Suggestions. — The  integral  part,  a  +  ar,  multiplied  by  the  denom- 
inator a—x,  gives  a''— .c^  From  this  subtracting  a^  +  2aiC4-a;',  as  the 
sign  before  the  fraction  indicates,  we  have  —2ax—2x^.     Hence  the 

,,  .    —2ax-2x' 

result  18 . 

a—x 

We  may  also  write  this  thus : 

a''  +  2ax  +  x^      a''—x''  —  {a''  +  2ax  +  x^ 
a—x  a—x 

Now  the  quantity  in  the  parenthesis  is  to  be  subtracted  from  the 
a'—x^.     Hence,  changing  the  signs  of  the  terras  in  the  parenthesis, 

//-         /y*^         ff'-^         0|i7/y         /f*^ 

and  dropping  the  marks,  we  have .      Adding 

.    .,  ,  .    ,  —2ax—2x^ 

similar  terms,  this  becomes . 

a—x 


4.  Reduce  1 5 — r^ —  to  the  form  of  a  fraction. 

a^-\-i^ 


Result, 


«2  +  ^ 


6.  Reduce  a—x to  the  form  of  a  fraction. 

a  +  x 

Result,  • 

a+x 

^ 5 

6.  Reduce  3x = —  to  the  form  of  a  fraction. 

bx 


Result,   — 

ox 


24  INTRODUCTION^. 


TO  REDUCE  FRACTIONS  TO  EQUIVALENT  FRAC- 
TIONS HAVING  A  COMMON  DENOMINATOR. 

30,  Give  the  rules  of  arithmetic  for  this  process. 

We  shall  use  only  the  method  of  multiplying  both  terms 
of  each  fraction  by  the  denominators  of  all  the  other 
fractions. 

Ex.  1.  Reduce  -,-,-  to  equivalent  fractions  having  a 
X  y  z 

nijz    i)ocz    cxu 
common  denominator.  Results,  -^ ,  —  ,  —^  • 

xyz    xyz    xyz 

2.  Reduce  7-  ?  —  >  t  to   equivalent  fractions  having  a 

common  denominator.     Also,  — ^ — , ,  ^' 

x  +  y'  x—y'  3 

Results  of  the  last,  ^^^33^ ,  ^^^ZT^f  '  ^^^ZT^f 


TO    ADD    FRACTIONS. 

31.  Repeat  the  rules  of  arithmetic  for  this  purpose. 


31a; 


Ex.  1.  Add  %,  %,  and  f-  Sum, 


2.  Add  |;  and  ^^.  Sum,'-^^ 

3.  Add  ^^  and  ^^  Sum,  ^^• 

4.  Add  ^-±\,  and  ^J.  Sum,  ^Jt^- 

5.  Add  \-^^,  and  i±^.  5«.m,  ^^. 


FRACTIONS    IN    THE    LITERAL    NOTATION.  25 

TO    SUBTRACT    FRACTIONS. 

32,   What  is  the  rule  given  in  arithmetic  for  this 
purpose. 

Ex.  1.  From  —  take  -r*  Rem.,  ^7^- 

rt         TT_  «     X     1  *  ^  2«  — 35 

2.  From  -  take  ^' 


3. 

From 

x—\ 
3 

take 

x  +  2 
5 

4. 
5. 

From 
From 

1 
x—y 

1-x^ 

take 
ta,ke 

1 
x-\-y 

1+x^ 

6 

Qfim 

2a:-ll 

15 

Rem. 

2y 

Rfim 

-^a? 

l-fa:2        ^  l—x^  -""'''  i_a4 


MULTIPLICATION  OP    FRACTIONS. 

33,  How    is    a   fraction  viultiplied    by    a  whof-e 
number  ? 

Ex.  1.  Multiply  ll  by  Sal  Prod.,  ^'. 

2.  MuUiply  3^^  by  a  +  J.  Pro^.,  gl|. 

3.  Multiply  \^^  by  l-o:.  Pro^.,  1^. 

4.  Multiply  g^  by  3a.  Prorf.,  ?|. 

5.  Multiply  3x  by  y-  P?W.,  -r-* 


26  INTRODUCTION. 


6.  Multiply  -y— ^f-7T9  by  a  +  ^.  Prod,, ^ — 

7.  Multiply  106^2  by  ^.  Prod.,  ?5^. 

8.  Multiply  I  by  6  ;   by  8 ;  by  10. 

9.  Multiply  ^  by  3  ;   ^^  by  5. 


34,  Oive  the  rule  for  multiplying  one  fraction  by 
another. 

Ex.  1.  Multiply  I  by  A.  Prod.,  g. 

2.  Multiply  J  by  -•  Pro(7.,  t^- 

!j  If 

3.  Multiply  — —  by  --^— •  Pro^Z., 


5a      ^  2— a  '  10a— 5a2 

4.  Multiply  — -!^-  by  -— •  Prori.,  ^ 

5                 7  7 

5.  Multiply  -j^    by  — j-  /'«rf..  -^-. 


DIVISION    OP   FKACTIONS. 
33,  How  is  a  fraction  divided  by  a  whole  number  f 

Ex.  1.  Divide  g  by  3x«.  Qmt,  ~ 

2.  Divide  ^  by  2a2j.  Quot.,  ,^- 


HOW  PROBLEMS  ARE  SOLVED  IN  ALGEBRA.     27 


3.  Divide by  a—o.  Quot., 


36,  How  is  any  quantity  divided  by  a  fraction  ? 
Ans.  By  midtiplying  it  by  the  divisor  inverted. 

Ex.  1.  Divide  6  by  %  Quof.,  -• 

2.  Divide  ^  by  -•  Quot.,  —  • 

3.  Divide—  by— ^.  0^^o^.,  -y-^- • 


4.  Divide  i  by 


Quot., 


ax-\-ay—x — y 


cm  +  dm — en — dn 

5.  Divide f-^-    by  -^.  Q^lot., ^. 

6.  Divide  — ^  by  ^,-^.  0^.o^  -3^. 


ION    IX. 


HOW    PROBLEMS   ARE   SOLVED    IN    ALGEBRA. 

Ex.  1.  John  is  3  times  as  old  as  James,  and  the  sum  of 
their  ages  is  32 ;  how  old  is  each  ? 


28  IKTRODUCTTON". 

Solution. — To  solve  this  by  Aritlimetic,  we  reason  thus:  Since 
John's  age  is  3  times  James's,  both  their  ages  together  make  4 
times  James's  age,  and  this  is  32  years.  Now  4  times  James's 
age  =  32  years.  Hence,  James's  age  is  i  of  32,  or  8  years ;  and 
John's  age  being  3  times  James's,  is  3  x  8,  or  24  years. 

To  solve  it  by  Algebra,  we  proceed  as  follows :  Let  x  represent 
James's  age ;  then,  since  John  is  3  times  as  old,  dx  will  represent 
his;  the  sum  of  their  ages  will  be  Bx  +  x.  Now  since  the  sum  of 
their  ages  is  32,  3.i'  +  x=S2;  or,  4a;  =  32.  If  4a;  =  32,  a;  =  ^  of  32, 
or  x  =  S.  Hence,  James's  age  is  8,  and  John's  being  3a;,  is  3  x  8, 
or  24. 

S7.  The  expression  3ic+a;  =  32  is  what  is  called  an 
Equation ;  and  it  is  by  the  use  of  equations  that  we 
solve  problems  in  Algebra. 

2.  A  merchant  said  that  he  had  72  yards  of  a  certain 
kind  of  cloth,  in  three  rolls.  In  the  first  roll,  there  were 
a  certain  mimber  of  yards  ;  in  the  second,  3  times  as  many 
as  in  the  first ;  and  in  the  third,  twice  as  many  as  in  the 
first.     How  many  yards  were  there  in  the  first  roll  ? 

Suggestions. 
The  equation  is  a?  +  3a;  +  2a;  =  72 

Now,  X  +  3a;  +  2a;  is  6a;,  hence  6a;  =  72 

And  if  6a;  =  72,  a-,  or  la;,  is  |  of  72,  a-  =  12 

Queries. — What  does  the  a;  stand  for?  Answer.  The  number  of 
yards  in  the  first  roll.  In  this  problem  which  is  the  most,  a;4-  3a;  +  2a;, 
or  72  ?  To  start  with,  do  you  know  how  much  x  is  ?  Then  is  it  a 
Jcnoicn.,  or  an  unknown  quantity,  at  the  outset  ? 


38.  The  answer  to  a  problem  is  represented  in  the 
beginning  of  the  solution  by  one  of  the  latter  letters  of 
the  alphabet,  usually  x,  if  there  is  need  of  but  one  letter, 
and  is  called  th(^  Unknown  Quantity. 

3.  A  boy  on  being  asked  how  old  he. was,  replied,  "If 
you  add  to  my  age  3  times  my  age,  and  5  times  my  age. 


HOW  PROBLEMS  ARE  SOLVED  IN  ALGEBRA.     29 

and  subtract  twice  my  age,  the  result  will   be  49  years. 
How  old  was  he  ? 

Suggestions. 
Tlie  equation  is  x  +  ^x-\-5x—2x  =  49 

Hence,  since  x  +  Sx-\-5x—2x  is  7x,  7a;  =  49 

If  7a;  =  49,  a;  =  -f  of  49,  or  aj  =  7 

4.  There  are  three  times  as  many  girls  as  boys  in  a 
p.irty  of  60  children.  How  many  boys  are  there?  How 
many  girls  ? 

5.  In  a  barrel  of  sugar  weighing  200  lbs.,  there  are  three, 
varieties.  A,  B,  and  0,  mixed.     There  is  7  times  as  much 
of  B  as  of  A,  and  twice  as  much  of  C  as  of  A.    How  much 
of  A  is  there  ?    How  much  of  each  of  the  other  kinds  ? 

Ans.,  Of  A,  20  lbs. ;  of  B,  140  lbs. ;  of  C,  40  lbs. 

6.  There  were  4  kinds  of  liquor  put  into  a  cask,  2  times 
as  much  of  the  second  as  of  the  first,  2  times  as  much  of 
the  third  as  of  the  second,  and  2  times  as  much  of  the 
fourth  as  of  the  third.  The  cask  sprang  a  leak,  and  three 
times  as  much  leaked  out  as  was  put  in  of  the  first  kind, 
when  it  was  found  that  there  were  36  gallons  remaining. 
How  much  was  there  put  in  of  each  kind  ? 

Suggestion.— The  equation  is        a!  +  2a;  +  4a;  +  8a;— 3aj  =  36. 


7.  In  a  pasture  there  are  a  certain  number  of  cows 
and  23  sheep,  in  all  34  animals.  How  many  cows  are 
there  ? 

Solution. — As  it  is  the  number  of  cows  we  seek,  let  x  represent 
the  numVjer  of  cows.  Then  ar  +  23  is  the  number  of  animals  in  the 
pasture,  and  the  equation  is 

aj+23  =  34. 

Now  the  ar  +  23  means  just  the  same  thing  as  the  34,  that  is 
x  +  2d  =  34.  So,  if  we  subtract  23  from  each,  there  will  be  just  as 
much  left  of  one  as  of  the  other.  Subtracting  23  from  a* +  23,  there 
remains  a-,  and  subtracting?  23  from  34,  there  remains  11.  Hence 
a;  =  11 ;  that  is,  there  are  11  cows. 


30  IKTEODITOTION". 

39,  The  part  of  an  equation  on  the  left  of  the  sign  =  is 
called  the  First  Member,  and  that  on  the  right,  the 
Second  Member. 

Note. — The  pupil  must  not  think  because  these  examples  are  so 
simple  that  he  can  "  do  them  in  his  head,"  that  therefore  algebra  is 
a  very  clumsy  method  of  solving  problems.  He  will  find  by  and 
by,  that  though  the  equation  does  not  really  help  any  in  the  solu- 
tion of  such  simple  questions,  it  will  solve  a  great  many  very  diffi- 
cult ones  about  which  he  might  puzzle  his  brains  a  great  while  to 
no  purpose,  if  algebra  did  not  come  to  his  aid.  Stick  to  it,  then, 
and  learn  how  to  use  this  new  instrument,  the  Equation,  and  you 
will  by  and  by  find  it  wonderfully  useful.  It  is  a  grand  patent  for 
solving  problems. 

8.  In  a  certain  pasture  there  are  three  times  as  many 
horses  as  cows,  and  20  sheep.  In  all  there  are  100  animals. 
How  many  cows  are  there  ?    How  many  horses  ? 

Suggestions. 
The  equation  is  a'4-3a;  +  20  =  100 

Subtracting  20  from  each  member,  sj  +  Sj;  =  80 

Uniting  the  terms  of  the  first  member,         4a?  =  80 
Dividing  each  member  by  4,  x  =  20 

Hence  there  are  20  cows ;  and,  as  there  are  three  times  as  many 
horses  as  cows,  there  are  60  horses. 

9.  In  a  .basket  of  60  apples  there  are  4  times  as  many 
red  apples  as  yellow,  and  10  green  apples.  How  many 
yellow  apples  are  there  ?    How  many  red  ? 

10.  John  and  James  together  have  75  cents.  James 
has  25  cents  less  than  John.     How  many  cents  has  John  ? 

Suggestions. — Let  x  represent  the  number  of  cents  which  John 
has.  Then,  as  James  has  25  cents  less,  ir— 25  will  represent  what 
he  has.     But  both  together  have  75  cents.     Hence  the  equation  is 

x-\-x—25  =  75. 

Now,  if  we  drop  the  —25  from  the  first  member,  we  make  this 
member  25  greater  than  it  now  is,  /.  e.,  x  +  x  is  25  greater  than 


HOW  PROBLEMS  ARE  SOLVED  IN  ALGEBRA.     31 

x  +  x—25.     Therefore,  if  we  add  25  to  the  second  member,  making 

it  100,  the  members  will  still  be  equal.* 

This  gives  x-\-x—  100 

or,  %x  =  100 

Hence  a?  =  50 

That  is,  John  has  50^. 

11.  A  merchant  has  90  yards  of  cloth  in  two  pieces. 
The  longer  piece  lacks  ten  yards  of  containing  3  times  as 
much  as  the  shorter.     How  much  in  each  piece  ? 

Suggestion. — The  equation  is  a;  +  3a!— 10  =  90. 

12.  Divide  the  number  50  into  two  parts  so  that  one 
part  shall  lack  10  of  being  5  times  the  other. 

Suggestions. — The  parts  are  represented  by  ar,  and  5a5— 10.  They 
are  10  and  40. 

13.  Divide  the  number  50  into  3  parts,  such  that  the 
second  shall  be  5  more,  and  the  third  15  less  than  the  first. 

Tlie  parts  are  20,  25,  and  5. 
Suggestion. — The  equatiou  is  a^  +  ir-f-5  +  iC— 15  =  50. 

14.  There  are  52  animals  in  a  field.  Twice  the  number 
of  cows +  11  is  the  number  of  sheep,  and  3  times  the  num- 
ber of  cows  —  13  is  the  number  of  horses.  How  many  of 
each  kind  ?  Ans.,  9  cows,  29  sheep,  and  14  horses. 


16.  A  man  said  of  his  age, 

'*  If  to  my  age  there  added  be 
One-half, t  one-third,  and  three  times  three, 
Six  score  and  ten  the  sum  will  be. 
What  is  my  age  ?     Pray  show  it  me.'* 

X       X 

Suggestion. — The  equation  is  x-k---\-  --\-%  =  130. 


*  The  word  "  Transposition  "  is  purposely  omitted  from  this  introdnction.  Nor 
la  the  idea  designed  to  be  presented.  It  will  ])e  better  for  the  pupil  to  "  think 
oat "  the  process,  as  above. 

t  Meaning  "  one-half  my  rtg-g,"  "  one-thinl  my  age."" 


32  INTRODUCTION. 

Subtracting  9  from  each  member, 

Now,  we  can  get  rid  of  the  fractions  in  the  first  member  by  mul- 
tiplying it  by  6,  the  product  of  both  the  denominators.  Thus,  6 
times  the  first  member  is  6a?  +  3aj  +  2aj.  Then,  if  we  also  multiply 
tl'c  seconil  member  by  6,  the  products  will  be  equal.  For  if  two 
quantities  are  equal,  6  times  one  of  them  is  equal  to  6  times  the 
other. 

Hence  we  have  ^x-^Zx-\-^x  =  726 

Uniting  terms,  \\x  =  726 

Dividing  by  11,  a;  =  66 

16.  Mary  gave  half  her  books  to  Jane,  and  one-third  of 
them  to  Helen,  when  she  had  but  2  left.  How  many  had 
she  at  first  ? 

Suggestions. — Let  x  represent  the  number  of  books  Mary  had  at 

first.     Then  she  gave  Jane  - ,  and  Helen  -  books.     And  what  she 

gave  the  other  girls,  added  to  what  she  had  left,  makes  all  she  had 
in  the  first  place.     Hence  the  equation  is 
x     X     ^ 

Multiplying  each  member  by  6,  3a;-f2aJ4-12  =  6aJ 

Subtracting  5a;  from  each  member,  12  =  « 

That  is,  Mary  had  12  books  at  first. 

17.  A  boy  lost  25  cents  of  some  money  which  his  uncle 
gave  him,  and  gave  half  he  had  left  to  his  brother.  He 
then  earned  50  cents,  when  he  had  just  as  much  as  his 
uncle  gave  him.    How  much  did  his  uncle  give  him  ? 

Suggestions. — Let  x  =  the  number  of  cents  his  uncle  gave  him. 

Then  he  had  ic— 25  cents  after  losing  25  cents.     After  giving  away 

2» 25 

half  of  this,  he  had  the  other  half,  or  — -—   cents,  left.     He  then 

2 

earned  50  cents,  and  the  amount  he  had  was  equal  to  what  his 

uncle  gave  him. 


HOW  PROBLEMS  ARE  SOLVED  IN  ALGEBRA.     33 

Hence  the  equation  is  -— —  +  50  =  a; 

Multiplying  each  member  by  3,  «— 25  + 100  =  2x 

Uniting,  —25  +  100  makes  75,  and  a;+75  =  2x 

Subtracting  x  from  each  member,  75  =  « 

18.  A  boy  being  asked  how  many  marbles  he  had,  said, 
"  If  I  had  five  more  than  I  have,  half  the  number  sub- 
tracted from  30  would  leave  twice  as  many  as  I  now  have." 
How  many  marbles  had  he  ? 

Suggestions. — Letting  x  represent  the  number  of  marbles  the  boy 
had,  the  equation  is 

Now  there  is  a  little  peculiarity  about  this  equation,  which  the 
pupil  must  be  careful  to  notice  whenever  it  occurs,  or  he  will  make 
a  great  many  mistakes.  It  is  this :  When  we  multiply  each  mem- 
ber by  2,  to  get  rid  of  the  fraction,  we  must  write  60— «— 5  =  4a;. 
The  mistake  would  be  to  write  60— d;  +  5  =  4a;.     The  explanation 

aj  +  5 
is,  that  the  —  sign  before  the  fraction  — --  does  not  belong  to  the 

a;,  but  to  the  fraction  as  a  whoU.    The  sign  of  x  in  the  fraction 

—  — —  is  + ,  since  when  no  sign  is  expressed  +  is  understood. 
2 

What  then  becomes  of  the  —  sign  before  the  fraction,  if  it  is  not  the 

same  as  the  sign  of  a;  in  the  equation  60— a;— 5  =  4a;?     It  has  been 

dropped,  since  the  thing  signified  by  it  has  been  performed,  and 

the  —  sign  before  the  x  is  the  sign  of  that  term  in  the  original 

equation,  changed.     In  like  manner  we  subtract   +5,  by  changing 

its  sign.  The  boy  had  11  marbles. 

19.  What  is    the  value  of    x    in    the    equation    '6x— 


Suggestions.— Multiplying  each  member  by  3, 

We  have  9a;— 2  +  2a;  =  64 

Hence,  a;  =  6 


34  IlsqTRODUCTIOlS'. 

20.  Find  the  value  of  x  in  the  equation 

x-\ ^r —  =  12 ~ Ans.,,  X  =  6. 

21.  What  is  the  value  of  x  in  the  equation 

x—1      X-^4:        ^„      ^+3^  .  .^. 

— — —  =  15 -r-  ?  Ans.f  X  =  40^. 

22.  Show  that  in 

ic— 1       ^     23—0;     4+a; 


23.  Two  boys  were  to  divide  32  marbles  between  them 
so  that  1^  of  what  one  had  should  be  5  less  than  what  the 
other  had.     How  many  was  each  to  have  ? 

Suggestions. — Letting  x  —  what  one  had, 

then  32— a;  =  what  the  other  had. 

The  equation  is  o  +  ^  =  32— a;, 


2 
32— a: 


+  5  =x. 


^uory. — Why  will  either  equation  answer  the  purpose? 

24.  What  number  is  that  to  which  if  7  be  added,  half 
the  sum  will  be  8  more  than  \  of  the  remainder  of  the 
number  after  3  has  been  subtracted  ? 

Equation  — — — -  =  8.  Ans.,  x  =  Id. 

25.  The  sum  of  two  numbers  is  sixteen,  and  the  less 
number  divided  by  three  is  equal  to  the  greater  divided  by 
five.     What  are  the  numbers  ? 

Suggestion. — Let  x  and  16— a;  represent  the  numbers. 

26.  Divide  twenty-two  dollars  between  A  and  B,  so  that 
if  one  dollar  be  taken  from  three-fourths  of  B's  share,  and 
three  dollars  be  added  to  one-half  of  A's  money,  the  sums 
shall  be  equal.     How  many  dollars  will  each  have  ? 


HOW  PROBLEMS  ARE  SOLVED  IN  ALGEBRA.     35 

27.  The  sum  of  two  numbers  is  thirty-three.  If  one- 
sixth  of  the  greater  be  subtracted  from  two-thirds  of  the 
less  number,  the  remainder  will  be  seven.  What  are  the 
numbers  ? 

28.  The  sum  of  A's  and  B's  money  is  thirty-six  dollars. 
If  five-eighths  of  B's,  less  two  dollars,  be  taken  from  three- 
fourths  of  A's,  the  difference  will  be  seven  dollars.  How 
many  dollars  has  each  ? 

29.  The  difference  between  two  numbers  is  twenty-five  ; 
and  if  twice  the  less  be  taken  from  three  times  the  greater, 
the  remainder  will  be  eighty.     What  are  the  numbers  ? 

30.  A  and  B  gain  money  in  trade,  but  A  receives  ten 
dollars  less  than  B.  If  A's  share  be  subtracted  from  twice 
B's,  the  remainder  wiU  be  fifty-seven  dollars.  How  much 
money  did  each  receive  ? 

31.  One  number  is  four  less  than  another,  and  if  twice 
the  less  be  subtracted  from  five  times  the  greater,  the 
remainder  will  be  thirty-eight.     What  are  the  numbers  ? 

32.  Two  farms  belong  to  A  and  B.  A  has  twenty  acres 
less  than  B.  If  twice  A's  number  of  acres  be  taken  from 
three  times  B's,  the  remainder  will  be  one  hundred.  How 
many  acres  has  each  ? 

33.  One  number  is  seven  less  than  another,  and  if  three, 
times  the  less  be  taken  from  four  times  the  greater,  the 
remainder  will  be  six  times  the  difference  between  the  two 
numbers.     What  are  the  numbers? 

34.  Anna  is  four  years  younger  than  Mary.  If  twice 
Anna's  age  be  taken  from  five  times  Mary's,  the  remainder 
will  be  thirty-five  years.    What  is  the  age  of  each  ? 

35.  One  number  is  ten  less  than  another.  If  three  times 
the  less  be  taken  from  five  times  the  greater,  the  remainder 
will  be  seven  times  the  difference  of  the  two  numbers. 
What  are  the  numbers  ? 


A  BRIEF  SURVEY  OF  THE  OBJECT  OF  PURE  MATHE- 
MATICS AND   OF   ITS   SEVERAL  BRANCHES. 

/.  Pure  Mathematics  is  a  general  term  applied  to 
several  branches  of  science,  which  have  for  their  object  the 
investigation  of  the  properties  and  relations  of  quantity 
— comprehending  number,  and  magnitude  as  the  result  of 
extension — and  of  form. 

2.  The  Several  Branches  of  Pure  Mathematics  are 
Arithmetic,  Algebra,  Calculus,  and  Geometry. 

5.  Arithmetic,  Algebra,  and  Calculus  treat  of  number, 
and  Geometry  treats  of  magnitude  as  the  result  of  extension. 

4.  Quantity  is  the  amount  or  extent  of  that  which  may 
be  measured ;  it  comprehends  number  and  magnitude. 

The  term  quantity  is  also  conventionally  Jipplied  to  sym- 
bols used  to  represent  quantity.  Thus  25,  m,  xi,  etc.,  are 
called  quantities,  although,  strictly  speaking,  they  are  only 
representatives  of  (juan titles. 

It  is  not  easy  to  give  a  philosophical  account  of  the  idea  or  idea.^, 
represented  by  the  word  Quautity  as  used  in  Mathematics;  antl 
doubtless,  different  persons  use  the  word  in  somewhat  different 
senses.  It  is  obviously  incorrect  to  say  that  "  Quantity  is  anything 
which  can  be  measured."     Quantity  may  be  affirmed  of  any  such 


2  INTKODUCTION. 

concept ;  nevertheless,  it  is  not  the  thing  itself,  but  rather  the 
amount  or  extent  of  it.  Thus,  a  load  of  wood,  or  a  piece  of  ground, 
can  be  measured ;  but  no  one  would  think  of  the  wood  or  piece  of 
ground  as  being  the  quantity.  The  quantity  (of  wood  or  ground) 
is  rather  the  am,ount.  oi  extent  of  it. 

The  word  is  very  convenient  as  a  general  term  for  mathematical 
concepts,  when  we  wish  to  speak  of  them  without  indicating 
whether  it  is  number  or  magnitude  that  is  meant.  Thus  we  say, 
"  m  represents  a  certain  quantity,"  and  do  not  care  to  be  more 
specific. 

As  applied  to  number,  perhaps  the  term  conveys  the  idea  of  the 
whole,  rather  than  of  that  whole  as  made  up  of  parts.  It  is,  there- 
fore, scarcely  proper  to  speak  of  multiplying  by  a  quantity ;  we 
should  say,  by  a  number.  On  the  other  hand,  when  we  apply  the 
term  quantity  to  magnitude,  it  is  with  the  idea  that  magnitude 
may  be  measured,  and  thus  expressed  in  number. 

The  distinction  between  quantity  and  number  is  marked  by  the 
questions,  ''  How  much  ?  "  and  "  How  many  ? " 

5.  Number  is  quantity  conceived  as  made  up  of  parts, 
and  answers  to  the  question,  *^How  many?" 

Thus,  a  distance  is  a  quantity ;  but,  if  we  call  that  distance  5, 
we  convert  the  notion  into  number,  by  indicating  that  the  distance 
under  consideration  is  made  up  of  parts.  Again,  m  may  mean  a 
value,  as  of  a  farm.  We  may  or  may  not  conceive  it  as  a  number 
(as  of  dollars).  If  we  think  of  it  simply  in  the  aggregate,  as  the 
worth  of  a  farm,  m  represents  quantity;  if  we  think  of  it  as  made 
up  of  parts  (as  of  dollars)  it  is  a  number. 

6*  Number  is  of  two  kinds.  Discontinuous  and  Con- 
tinuous. 

7.  Discontinuous  Number  is  number  conceived  as 
made  up  of  finite  parts  ;  or  it  is  number  which  passes  from 
one  state  of  value  to  another  by  the  successive  additions 
or  subtractions  of  finite  units ;  i.  e.,  units  of  appreciable 
magnitude. 


THE    OBJECT    OF    PC7RE    MATHEMATICS.  3 

8,  Continuous  Number  is  number  which  is  conceived 
as  composed  of  intinitesimal  parts ;  or  it  is  number  which 
passes  from  one  state  of  value  to  another  by  passing 
through  all  intermediate  values,  or  states. 

Number,  as  considered  in  Arithmetic,  and  in  this  volume,  is 
Discontinuous  Number.  Thus  5  grows  till  it  becomes  9,  by  taking 
on  additions  of  units  of  some  conceivable  value ;  as  when  we  consider 
it  as  passing  from  5  to  9,  thus,  5,  5  + 1  or  6,  6  + 1  or  7,  7  + 1  or  8, 
8  4- 1  or  9.  If  the  increment  were  any  fraction,  however  small,  t/ie 
fonn  of  the  cojiceptit/n  would  he  the  same. 

Time  affords  a  good  illustration  of  Continuous  Number.  We 
usually  conceive  time  as  a  discontinuous  numher^  as  when  we  think 
of  it  as  made  up  of  hours,  days,  weeks,  etc.  But  it  is  easy  to  see 
that  such  is  not  the  way  in  which  time  actually  grows.  A  period 
of  one  day  does  not  grow  to  be  a  period  of  one  week  by  taking  on 
a  whole  day  at  a  time,  or  a  whole  hour,  or  even  a  whole  second. 
It  grows  by  imperceptible  increments  (additions).  These  incon- 
ceivably small  parts,  by  which  time  is  actually  made  up,  we  call 
infinitesimals;  and  number,  when  conceived  as  made  up  of  such 
infinitesimals,  we  call  Continuous  Number. 

9,  Arithmetic  treats  of  Discontinuous  Number,— 

of  its   nature  and  properties,  of  the  various  methods  of 
combining  and  resolving  it,  and  of  its  application  to  prac 
tical  affairs. 

The  leading  topics  of  Arithmetic  are : 

1.  Notation;  i.  e.,  methods  of  representing  number,  as  by  the 
characters,  1,  2,  3,  4,  etc.,  or  by  letters,  as,  a,  5,  m,  n,  x,  y,  etc. 

2.  Properties  of  Numbers  or  deductions  from  the  methods  of 
Notation. 

3.  Reduction,  as  from  one  scale  to  another,  from  one  denomina- 
tion to  another,  from  one  fractional  form  to  another,  or,  in  short, 
from  any  one  form  of  expression  to  another  equivalent  form. 

4.  The  various  methods  of  combining  number,  as  by  addition, 
multiplication,  and  involution. 

5.  Resolving  Number,  as  by  subtraction,  division,  and  evo- 
lution. 


4  INTRODUCTION^. 

And  all  these  processes  as  affected  by  the  use  of  any  notation, 
and  upon  integral  or  fractional  discontinuous  numbers  of  any 
kind. 

Arithmetic,  therefore,  philosophically  considered,  embraces 
much  that  is  usually  classed  as  Algebra.  Thus  all  that  usually 
precedes  Simple  Equations,  and  all  that  is  embraced  in  this  Part  I. 
is  simply  a  repetition  and  extension  of  the  processes  of  Arithmetic 
with  a  new  notation — the  literal.  Again,  logarithms  are  nothing 
but  a  new  scheme  of  notation,  by  means  of  which  certain  com- 
binations are  more  readily  effected ;  and  the  making  of  logarithms 
is  but  a  reduction  from  one  form  of  expression  to  an  equivalent 
one  in  another  notation.  In  the  ordinary  notation,  a  certain 
number  is  represented  thus,  256 ;  in  the  logarithmic  notation  it  is 
2.40834. 

10,  Algebra  treats  of  the  Equation,  and  is  chiefly 
occupied  in  explaining  its  nature  and  the  methods  of 
transforming  and  reducing  it,  and  in  exhibiting  the 
manner  of  using  it  as  an  instrument  for  mathematical 
investigation. 

The  whole  province  of  the  relations  of  quantity,  continuous  or 
discontinuous  number,  is  covered  by  Algebra,  so  far  as  the  equation 
can  be  made  the  instrument  of  investigation.  Much,  therefore,  of 
what  is  found  in  our  Arithmetics  can  be  more  expeditiously  treated 
by  Algebra.  Such  are  the  subjects  of  Ratio,  Proportion,  the 
Progressions,  Percentage,  Alligation,  etc.  In  fact^  the  equation  is 
the  grand  indrimient  of  mathematical  investigation^  and  demonstrates 
its  efficiency  in  every  depa.rtment  of  the  science.  To  hope  to  get  on  in 
mathematics  without  Algebra,  is  to  expect  to  walk  without  feet. 

11,  Calculus  treats  of  Continuous  Number,  and  is 
chiefly  occupied  in  deducing  the  relations  of  the  infini- 
tesimal elements  of  such  number  from  given  relations 
between  finite  values,  and  the  converse  process,  and  also 
in  pointing  out  the  nature  of  such  infinitesimals  and  the 
methods  of  using  them  in  mathematical  investigation. 

12,  G-eometry  treats  of  magnitude  and  form  as  the 
result  of  extension  and  position. 


tX)OICO-MATHEMATlCAL    TERMS.  5 

The.principal  divisions  of  the  science  of  Geometry  are : 

1.  The  Ancient,  Special,  or  direct  Geometry  (the  common  Geome- 
try of  our  schools),  including  Trigonometry,  Conic  Sections,  and  all 
other  geometrical  inquiries  conducted  upon  these  methods. 

2.  The  Modern,  Indirect,  or  General  Geometry  (usually  called 
Analytical),  and 

3.  Descriptive  Geometry. 

SYNOPSIS. 

Subject  of  Section.  Quantity :  Uses  of  the  term. 

Definition  of  Pure  Mathematics:  Number:    Illustration  of :   Kinds 

Several  Branches  of.  of:  Illustration  of  kinds. 

Subject  Matter  of  the  Several  Arithmetic :  Topics  of. 

Branches.  Algebra— Calculus— Geometry. 


mon  II 


LOGICO-MATHEMATICAL    TERMS. 

13.  A  Proposition  is  a  statement  of  something  to  be 
considered  or  done. 

Illustration.— Thus,  the  common  statement,  "Life  is  short,"  is  a 
proposition ;  so,  also,  we  make,  or  state  a  proposition,  when  we 
say,  "Let  us  seek  earnestly  after  truth." — "The  product  of  the 
divisor  and  quotient,  plus  the  remainder,  equals  the  dividend,"  and 
the  requirement,  "  To  reduce  a  fraction  to  its  lowest  terms,"  are 
examples  of  Arithmetical  propositions. 

14.  Propositions  are  distinguished  as  Axioms,  Theorems, 
Lemmas,  Corollaries,  Postulates,  and  Problems. 

15.  An  Axiom  is  a  proposition  which  states  a  princi- 
ple that  is  so  simple,  elementary,  and  evident,  as  to  require 
no  proof. 

Illustration. — Thus,  "A  part  of  a  thing  is  less  than  the  whole  ot 
it,"  "  Equimultiples  of  equals  are  equal,"  are  examples  of  axioms. 


6  INTRODUCTION. 

If  any  one  does  not  admit  the  truth  of  axioms,  when  he  understands 
the  terms  used,  we  say  that  his  mind  is  not  sound,  and  that  we 
cannot  reason  with  him. 

16,  A  Theorem  is  a  proposition  which  states  a  real  or 
supposed  fact,  whose  truth  or  falsity  we  are  to  determine 
by  reasoning. 

Illustration.— "If  the  same  quantity  be  added  to  both  numerator 
and  denominator  of  a  proper  fraction,  the  value  of  the  fraction  will 
be  increased,"  is  a  theorem. 

17 »  A.  Demonstration  is  the  course  of  reasoning  by 
means  of  which  the  truth  or  falsity  of  a  theorem  is  made 
to  appear.  The  term  is  also  applied  to  a  logical  statement 
of  the  reasons  for  the  processes  of  a  rule.  A  solution  tells 
how  a  thing  is  done ;  a  demonstration  tells  why  it  is  so  done. 
A  demonstration  is  often  called  proof. 

IS.  A  Lemma  is  a  tlieorem  demonstrated  for  the 
purpose  of  using  it  in  the  demonstration  of  another 
theorem. 

Illustration. — Thus,  in  order  to  demonstrate  the  rule  for  finding 
the  greatest  common  divisor  of  two  or  more  numbers,  it  may  be 
best  first  to  prove  that  "  A  divisor  of  two  numbers  is  a  divisor  of 
their  sum,  and  also  of  their  difference,"  This  theorem,  when 
proved  for  such  a  purpose,  is  called  a  Lemma. 

The  term  Lemmxi  is  not  much  used,  and  is  not  very  important, 
since  most  theorems,  once  proved,  become  in  turn  auxiliary  to  the 
proof  of  others,  and  hence  might  be  called  lemmas. 

10,  A  Corollary  is  a  subordinate  theorem  which  is 
suggested,  or  the  truth  of  which  is  made  evident,  in  the 
course  of  the  demonstration  of  a  more  general  theorem, 
or  which  is  a  direct  inference  from  a  proposition. 

Illustration. — Thus,  by  the  discussion  of  the  ordinary  process  of 
performing  subtraction  in  Arithmetic,  the  following  Corollary  might 
be  suggested  :  "  Subtraction  may  also  be  performed  by  addition,  as 
we  can  readily  observe  whnt  number  must  be  added  to  the  subtra- 
hend to  produce  the  minuend." 


LOGICO-MATHEMATICAL    TERMS.  7 

20,  A  Postulate  is  ii  proposition  which  states  that 
somethiug  can  be  done,  and  which  is  so  evidently  true 
as'  to  require  no  process  of  reasoning  to  show  that  it  is 
possible  to  be  done.  We  may  or  may  not  know  how  to 
perform  the  operation. 

Illustration. — Quantities  of  the  same  kind  can  be  added  together. 

21,  A  Problem  is  a  proposition  to  do  some  specified 
thing,  and  is  stated  with  reference  to  developing  the 
method  of  doing  it. 

Illustration. — A  problem  is  often  stated  as  an  incomplete  sentence, 
as,  "  To  reduce  fractions  to  forms  having  a  common  denominator." 

22,  A  Rule  is  a  formal  statement  of  the  method  of 
solving  a  general  problem,  and  is  designed  for  practical 
application  in  solving  special  examples  of  the  same  class. 
Of  course  a  rule  requires  a  demonstration. 

23,  A  Solution  is  the  process  of  performing  a  problem 
or  an  example.  It  should  usually  be  accompanied  by  a 
demonstration  of  the  process. 

24,  A  Scholium  is  a  remark  made  at  the  close  of  a 
discussion,  and  designed  to  call  attention  to  some  particular 
feature  or  features  of  it. 

Illustration. — Thus,  after  having  discussed  the  subject  of  multi- 
plication and  division  in  Arithmetic,  the  remark  that  "Division  is 
the  converse  of  multiplication,"  is  a  scholium. 


SYNOPSIS. 

Subject  of  Section. 

Lemma.    III. — Why  the  term  is 

Proposition.    /^/.— Varieties  of. 

unimportant. 

Axiom.     lU. — One  who  will  not 

Corollary.     III. 

admit  the  truth  of. 

Postulate.     III. 

Theorem.    lU. 

Problem.    How  stated.    III. 

Demonstration.      Difference   be- 

Rule. 

tween  a  solution  and  a  demon- 

Solution. 

stration. 

Scholium.    lU. 

PART     I 


jl^^Anjuui:  -^^^^^'^^^^^  larLfmJX'^/^  ^ 


FUNDAMENTAL     RULES. 


Section  ». 


^'— w- 


^atio^^ 


25,  A  System  of  Notation  is  a  system  of  symbols 
by  means  of  which  quantities,  the  relations  between  them, 
and  the  operations  to  be  performed  upon  them,  can  be 
more  concisely  represented  than  by  the  use  of  words. 

SYMBOLS    OF    QUANTITY. 
26»  In  Arithmetic,  as  usually  studied,  numbers  are  rep- 

NOTE.—Part  First  treats  of  the  familiar  operations  of  Addition, 
Subtraction,  Multiplication ,  Division,  Involution  a»d,  Evolution,  and 
the  tlieor//  of  Fractions,  The  only  difference  hettveen  the  jn'occsses  here 
developed  and  those  tvith  which  the  2)Hpil  is  already  familiar,  grotvs 
out  of  the  notatittn.  Hence  appears  the  appositeness  of  the  term 
Literal  Arithmetic.  Hence,  also,  the  teacher  should  he  careful  that 
the  pupil  see  the  unity  of  purpose,  and  the  reason  for  any  difference 
in  tnethod  of  execution. 


NOTATION.  9 

resented  by   the  characters,   1,  2,  3,  4,  5,  6,  7,  8,  9,  0, 
called  Arabic  figures,  or,  simply,  figures. 

27,  In  other  departments  of  mathematics  than  Arith- 
metic, numbers  or  quantities  are  more  frequently  repre- 
sented by  the  common  letters  of  the  alphabet,  a,  b,  c, 
.  ,  .  m,  n  .  ,  .  X,  y,  z.  These  letters  may,  however,  bo 
used  in  Arithmetic ;  and  the  Ai*abic  figures  are  used  in  all 
departments  of  mathematics. 

This  method  of  representing  quantities  by  letters  is 
often  called  the  Algebraic  method,  and  the  method  by 
the  Arabic  characters,  the  Arithmetical.  It  would  be 
better  to  call  the  former  the  Literal  method,  and  the 
latter  the  Decimal. 

28.  The  Literal  Notation  has  some  very  great  advan- 
tages over  the  decimal  for  purposes  of  mathematical 
reasoning : 

1st.  The  symbols  are  more  general  in  their  significa- 
tion; and 

2d.  We  are  enabled  to  detect  the  same  quantity  any- 
where in  the  process,  and  even  in  the  result.  Thus  it 
happens  that  the  processes  become  general  formulas,  or 
rules,  instead  of  special  solutions. 

Illustration. — To  illustrate  the  first  statement,  suppose  we  say  a 
boy  has  7  apples  ;  you  know  just  hoic  many  are  meant.  But  wb?  \ 
we  Bay  a  boy  has  b  apples,  nobody  can  tell  how  many  he  has.  In 
fact,  it  is  not  designed  to  tell  the  exact  number,  but  only  to  say 
that  he  has  some  number.  Again,  7  represents  the  same  number  of 
units  always;  but  a  letter  may  be  used  to  represent  any  number  of 
units  we  please;  or,  it  may  be  used,  as  we  have  just  said,  without 
our  caring  to  specify  any  precise  number  of  units.  This  may  seem 
to  be  a  very  unsatisfactory  kind  of  notation;  but  with  patience  its 
advantages  will  appear.  The  following  examples  will  illustrate 
the  general,  or  comprehensive  character  of  the  literal  notation 
more  fully. 


10  FUNDAMENTAL    RULES. 


EXAMPLES. 

Ex.  1.  A  boy  has  8  marbles  which  he  sells  for  three  cents 
each,  and  takes  his  pay  in  pencils  at  6  cents  each.  How 
many  pencils  does  he  receive  ? 

Suppose    we    answer    that    he    receives    3    times    8    divided 

3x8 
by    6,    or    — - —    pencils,    without    giving    the    number    more 
6 

explicitly. 

Now  take  a  similar  example,  nsing  the  literal  nota- 
tion ;  thus,  A  boy  sells  a  certain  number  of  marbles 
which  we  will  represent  by  c,  for  a  number  of  cents 
each,  which  we  will  call  m,  and  takes  his  pay  in  pen- 
cils at  l  cents  each.  How  many  pencils  does  he 
receive  ? 

We  will   answer  as  before,   and   say  he   receives  c  times  w, 

divided  by  5,  or  — —  pencils. 

The  pupil  will  notice  this  diflference  between  the  answers ; 
both,  as  they  now  stand,  simply  tell  what  operations  to  perform  in 
order  to  get  the  answers;  but,  in  the  former  case,  we  can  perform 
the  operations  and  get  the  explicit  answer,  4,  while  in  the  latter 
case,  we  can  only  leave  it  as  it  is. 

Sucn  an  answer  as  — r^ —  may  seem  to  the  pupil  to  be  no  answer 

0 

at  all ;  and  indeed  it  is  not  an  answer  in  the  same  sense  as  he  has 
been  accustomed  to  think  of  answers ;  nevertheless  it  is  often  more 
useful.      Notice  that  the  answer  4  is  only  true  for  the  specific 

example,  while  the  answer  — = —  is  true  in  every  like  example. 

We  also  observe  that  the  quantities  3,  8,  and  6  do  not  appear 
distinctly  in  the  numerical  answer,  4 ;  but  the  c,  m^  and  h  do  in  the 
literal,  and  would,  in  general,  however  complicated  the  problem. 
The  literal  answer  is  equivalent  to  the  rule.  Multiply  the  price 
of  one  7narUe  lyy  the  number  of  marlles^  and  divide  the  product  hy  the 
price  of  a  pencil. 


KOTATION.  11 

2.  One  boy  sold  5  pears  at  3  cents  each ;  another 
sold  6  apples  at  2  cents  each ;  and  a  third  sold  3  melons 
at  8  cents  each.    How  much  did  they  all  receive  ? 

Atis.,  51  cents. 

3.  One  boy  sold  b  pears  at  c  cents  each  ;  another  sold 
m  apples  at  n  cents  each  ;  a  third  sold  d  melons  at  g  cents 
each.     How  much  did  they  all  receive  ? 

Ans.,  bxc-\-mx7i-{-dxg  cents. 

Suggestions. — Notice  that  in  the  3d  example  the  several  quanti- 
ties of  the  problem  are  distinctly  seen  in  the  answer,  but  not  so  in 
the  answer  to  Mc.  2.  Moreover,  the  answer  to  Ex.  3  is  equally  true 
for  any  and  all  values  of  &,  c,  m,  n,  d,  and  g.  Consider  in  like 
manner  the  two  following : 

4.  If  I  buy  5  cords  of  wood  at  4  dollars  per  cord,  and 
pay  for  it  in  cloth  at  2  dollars  per  yard,  how  many  yards 
are  required  ?  Ans.,  10  yai'ds. 

6.  If  I  buy  a  cords  of  wood  at  b  dollars  per  cord,  and 

pay  for  it  in  cloth  at  c  dollars  per  yard,  how  many  yards 

are  required  ?  .       axb       ^ 

Ans., yards. 

6.  A  man  had  a  flock  of  m  sheep.  He  lost  2n  of  them 
and  raised  10a.  After  which  he  sold  the  flock  at  $c 
per  head,  taking  his  payment  in  cloth  at  $b  per  yard. 
What  operations  must  be  performed  on  these  numbers  in 
order  to  ascertain  the  number  of  yards  of  cloth  received  ? 
And  how  will  this  number  be  represented  ? 

7.  A  man  bought  3  horses.  He  gave  for  the  first  twice 
as  much  as  for  the  second,  and  for  the  third  c  times  as 
much  as  for  both  the  others.  If  x  represents  the  price 
of  the  first,  how  much  did  he  give  for  the  third  ? 

Ans.,  (•^-fo)  ^- 


12  FUNDAMENTAL    RULES. 

20,  In  using  the  decimal  notation  certain  laws  are 
established  in  accordance  with  which  all  numbers  can 
be  represented  by  the  ten  figures.  Thus,  it  is  agreed 
that  when  several  figures  stand  together  without  any 
other  mark,  as  435,  the  right  hand  figure  shall  signify 
units,  the  second  to  the  left,  tens,  the  third,  hundreds, 
etc.;  also  that  the  sum  of  the  several  values  shall  be 
taken.  This  number  is,  therefore,  4  hundreds  +  3  tens 
-}-  5  (units). 

30,  In  like  manner,  certain  laws  are  observed  in  repre- 
senting numbers  by  letters. 

FIRST    LAW. 

Known  Quantities,  that  is  such  as  are  given  in  a 
problem,  are  represented  by  letters  taken  from  the  first 
part  of  the  alphabet;  while  Unknown  Quantities,  or 
quantities  whose  values  are  to  be  found,  are  represented 
by  letters  taken  from  the  latter  part  of  the  al]3habet. 

Illustration.— A  grocer  has  two  kinds  of  tea,  one  of  which  is 
worth  a  cents  (any  given  number  being  meant  by  a)  per  pound, 
and  the  other  h  cents.  How  many  pounds  of  each  must  he  take  to 
make  a  chest  of  c  pounds,  which  will  be  worth  d  dollars  ?  In 
this  problem,  a,  &,  c,  and  d  are  the  given  or  known  quantities, 
and  hence  are  represented  by  letters  from  the  first  part  of  the 
alphabet.  The  unknown  or  required  quantities  are  the  nuraber  of 
pounds  of  each  of  the  two  kinds  of  tea.  We  therefore  represent 
the  number  of  pounds  of  the  first  kind  by  «,  and  of  the  second 
kind  by  y. 

Scholium. — This  law  is  not  very  rigidly  adhered  to,  except  that 
letters  after  and  including  t),  are  generally  used  to  represent  un- 
known quantities,  while  the  others  are  used  for  known  quantities. 
But  it  is  sometimes  convenient  to  use  a  different  notation.  Thus, 
in  problems  in  Interest,  the  principal  may  be  represented  by  j^, 
whether  it  is  known  or  unknown,  the  interest,  in  like  manner,  by  i, 
the  rate  per  cent,  by  r,  the  time  by  i,  etc. 


NOTATION.  18 

Accented  letters,  as  «',  «",  a",  «"", etc.,  (read  "«  prime," 
"  a  second,"  "  a  third,"  etc.),  and  letters  with  subscripts,  as 
fli,  fljj  «3,  cL^i  etc.,  (read  "a  sub  1,"  "a  sub  two,"  etc.),  are 
sometimes  used.  This  form  of  notation  is  used  when  there 
are  several  like  quantities  in  the  same  problem,  but  which 
have  different  numerical  values.  Thus,  in  a  problem  in 
which  several  walls  of  different  heights,  breadths,  and 
lengths,  are  considered,  we  may  represent  the  several 
heights  by  a',  a",  a",  etc.,  or  «, ,  a^,  a^,  etc.;  the  thick- 
nesses by  b',  i",  V" ,  etc.,  or  J,,  h<i,  J3,  etc.,  and  the  lengths 
by  r,  r',  Z'",  etc.,  or  U,U,k,  etc. 

The  Greek  letters  are  also  often  used  both  for  known 
and  unknown  quantities. 

The  student  will  notice  a  difference  between  Algebra  and 
Arithmetic,  in  that,  in  Algebra,  the  unknown  quantities 
(what  he  has  called  the  ** Answers"  in  Arithmetic)  are 
represented  in  solving  a  problem,  and  are  used  in  the 
solution  just  like  known  quantities.  This  device  gives 
Algebra  a  great  advantage  over  Arithmetic. 

SECOND     LAW. 

When  letters  are  written  in  connection,  without  any 
sign  between  them,  their  product  is  signified.  Thus  abc 
signifies  that  the  three  numbers  represented  by  a,  b,  and  c 
are  to  be  multiplied  together. 

Scholium  I— There  is  here  an  interesting  difference  between 
this  notation  and  the  decimal.  There  are  two  points  of  difference ; 
1st,  The  place  of  a  letter,  as  at  the  right  or  left,  has  nothing  to  do 
with  its  value ;  and  2d,  The  sign  understood  between  them  is  that 
of  multiplication  instead  of  addition,  as  in  the  decimal  notation.  In 
the  decimal  system,  if  we  write  the  three  figures  5, 4,  and  3,  as  we  have 
written  the  letters  a,  &,  and  c,  thus  543;  1st,  Each  figure  has  a  par- 
ticular value  dependent  upon  the  place  it  occupies,  the  5  repre- 
senting hundreds,  the  4  tens,  and  the  3  units ;   2d,  The  amount 


14  FUNDAMEITTAL    RULES. 

represented  is  500  +  40  +  3.  Moreover,  the  several  letters  are  not 
at  all  restricted  in  signification ;  a  may  represent  5,  48,  10.06,  or 
any  other  number,  however  small  or  however  large,  and  integral, 
fractional,  or  mixed ;  and  the  same  is  true  of  any  other  letter.  In 
fact,  the  meaning  usually  is,  that  a  represents  any  numljer,  ft  any 
other^  and  c  any  other,  etc.  The  same  letter,  however,  means  the 
same  thing  throughout  one  problem.  Such  expressions  as  oftc, 
mnxy^  etc.,  are  read  by  simjjly  naming  the  letters  in  order. 

Scholium  2. — When  figures  are  written  in  connection  with  let- 
ters, their  relation  to  the  letters  is  the  same  as  that  of  the  letters 
to  each  other.  Thus  4a6  means  the  continued  product  of  4,  a,  and  h. 
Also  125ai2/  means  the  continued  product  of  125,  «,  and  y. 

31,  K  character  like  a  figure  8  placed  horizontally,  oo , 
is  used  to  represent  what  is  called  Infinity,  or  a  quantity 
larger  than  any  assignable  quantity. 

Scholium- — By  an  infinite  quantity  is  not  meant  one  larger  than 
any  other,  or  the  largest  possible  quantity.  It  simply  means  a 
quantity  larger  than  any  assignable  quantity ;  ^.  e.^  larger  than  any 
one  which  has  limits.  Thus,  a  series  of  Is,  as  lit,  etc.,  repeated 
without  stopping,  represents  an  infinite  quantity,  because,  from  the 
method  of  conceiving  the  quantity,  it  is  necessarily  greater  than 
any  quantity  which  we  can  assign  or  mention.  If  we  assign  a  row 
of  9s  reaching  around  the  world,  though  it  is  an  inconceivably 
great  number,  it  is  not  as  great  as  a  series  of  Is  extending  without 
limit.  Moreover,  one  infinite  may  be  larger  than  another ;  for  a 
series  of  2s  extending  without  limit,  as  2  2  2  2,  etc.,  is  twice  as  large 
as  a  series  of  Is  conceived  in  the  same  way.  It  is  never  of  any  use 
to  try  to  comprehend  the  magnitude  of  an  infinite  quantity ;  we 
can  not  do  it;  although  we  can  compare  infinites  just  as  well  as 
finites. 

SYMBOLS    OP    OPERATION. 

32,  The  Symbols  of  Operation  used  in  Algebra 
are  the  same  as  in  Arithmetic,  or  in  any  other  branch  of 
mathematics ;  but,  to  refresh  the  memory,  we  will  repeat 
them. 

33,  The  perpendicular  cross,  + ,  is  called  the  plus  sign, 
and  read  *'  plus."    It  signifies  that  the  quantities  between 


NOTATIOX.  15 

which  it  is  placed  are  added  together.     Thus,  a-{-2cb-{-xm 
is  read,  "  a  plus  2cb  plus  xm.'' 

34,  A  short  horizontal  line,  — ,  is  called  the  minus  sign, 
and  is  read  "  minus."     It  signifies  that  the  quantity  before 
which  it  is  placed  is  to  be  subti*acted.    Thus,  a^2cb—xm 
^-VZax  is  read,  "  a  minus  'Zch  minus  xm  plus  12  «a;." 

33.  An  S-shaped  symbol  placed  horizontally,  '^,  is  some- 
times used  to  signify  the  difference  between  two  quantities. 
tt'^h  is  read,  "the  difference  between  a  and  ^."  This 
sign  differs  from  the  preceding  in  that  it  does  not  indicate 
which  of  the  two  quantities  is  to  be  taken  as  the  subtra- 
hend, while  the  minus  sign  requires  as  to  consider  the 
quantity  before  which  it  is  placed  as  the  subtrahend. 

30.  The  oblique  cross,  x,  and  a  simple  dot,  •,  are  each 
signs  of  muUiplication.  In  the  case  of  literal  factors,  the 
sign  is  usually  omitted,  according  to  the  second  law  of 
notation.  Thus,  4xaxc,  4-a.c,  and  iac,  signify  exactly 
the  same  thing. 

57.  The  signs  of  division  are,  a  horizontal  line  between 
two  dots,  having  the  dividend  at  the  left  and  the  divisor  at 
the  right,  as  12ac-^2b ;  or  the  dots  without  the  line,  as 
I2ac  :2b;  or  the  line  without  the  dots,  the  dividend  being 

written  above  and  the  divisor  below  it,  as  -^rr-  ;  each  of  which 

is  read,  "  12ac  divided  by  2b."  In  performing  division, 
the  divisor  is  sometimes  written  at  the  left  of  the  dividend 
and  separated  from  it  by  a  curved  line  ;  the  quotient  is 
then  written  at  the  right  and  separated  from  the  dividend  in 
the  same  manner:  as,  2a)12ac{QCy  in  which  2a  is  the  divisor, 
12ac  tlie  dividend,  and  6c  the  quotient.  Sometimes,  espe- 
cially in  Algebra,  the  divisor  is  written  on  the  right  of  the 
dividend,  and  separated  from  it  by  a  vertical  line,  the  quo- 


16  FUNDAMENTAL    RULES. 

tient  in  this  case  being  written  under  the  divisor.     Thus, 
12ac  \2a  is  the  last  example  above,  expressed  in  a  differ- 

6c 
ent  form. 

Scholium.— It  is  very  inelegant,  though  quite  common  in  some 

parts  of  the  country,  to  read  such  expressions  as  — ^,  "  12ac  over 

2o 

2&."     We  should  read  "  12ac  divided  by  26/' 

[Note. — Let  great  care  be  taken  that  the  nature  of  exponents,  as 
explained  in  the  succeeding  articles,  be  clearly  comprehended.  No 
little  difficulty  arises  from  an  imperfect  understanding  of  this  nota- 
tion. A  very  common,  though  very  erroneous  method  of  reading 
such  expressions,  greatly  aggravates  the  diffculty.] 

38.  A  Power  of  a  number  is  the  product  which  arises 
from  multiplying  the  number  by  itself,  i.  e.,  taking  it  a 
certain  number  of  times  as  a  %ctor. 

39.  A.  Root  of  a  number  is  one  of  several  equal  factors 
into  which  the  number  is  to  be  resolved. 

40.  An  Exponent  is  a  small  figure,  letter  or  other 
symbol  of  number  written  at  the  right  and  a  little  above 
another  figure,  letter  or  symbol  of  number. 

41.  A  Positive  Integral  Exponent  signifies  that  the 
number  affected  by  it  is  to  be  taken  as  a  factor  as  many 
times  as  there  are  units  in  the  exponent.  It  is  a  kind  of 
symbol  of  multiplication. 

Illustration.  2^  (read,  "  2,  third  power"),  signifies  that  two  is  to 
be  taken  three  times  as  a  factor,  /.  e.,  2x2x2,  or  8.  3*  (read,  '*  3, 
fourth  power"),  signifies  3  x  3  x  3  x  3,  or  81 ;  81  is  the  fourth  power 
of  3,  because  it  is  the  product  of  3  taken  four  times  as  a  factor,  a' 
is  aaaaa.  af'\  read  "  x,  mih  power,"  or  "  «,  exponent  w,"  is  xxx  .  . . 
etc.,  till  m  factors  ofx  are  taken. 

42.  A  Positive  Fractional  Exponent  indicates  a 
power  of  a  root,  or  a  root  of  a  power.     The  denominator 


NOTATION.  17 

specifies  the  root,  and  the  numerator  the  power  of  the 
number  to  which  the  exponent  is  attached. 

Illustration.  8^  (read  "  8,  exponent  f ,"  not  "  8,  |  power,"  there 
is  no  such  thing  as  a  |  power),  is  the  second  power  of  the  third  root 
of  8.     The  third  root  of  8  being  2,  and  tlie  second  power  of  2  being 

4,  8^  —  4.     We  may  also  understand  the  power  to  be  taken  first, 

and  then  the  root,  as  will  be  demonstrated  hereafter.  Thus,  8J 
is  the  third  root  of  the  second  power  of  8.  The  second  jjower  of  8 
is  64,  the  third  root  of  which  is  4,  which  is  the  same  result  as  was 

obtained  by  taking  the  root  first,  and  then  the  power.     (125)^  is  5. 

(125)^  is  25.  (32)f  is  8.  aj»  (read  "a;,  exponent  in  divided  by  n") 
means  that  x  is  to  be  resolved  into  n  equal  factors,  and  the  product 
of  m  such  factors  taken. 

43.  A  Negative  Exponent,  either  integral  or  frac- 
tional, signifies  the  reciprocal  of  what  the  expression  would 
be  if  the  exponent  were  positive. 

Illustration.      3-*  (read  "3,  exponent  —4")  signifies  5-4,01—- 

o  81 


-,  >  or   - .    x'"  18  —  ,  etc.     Also  8      is  -— ,  or  -  .     w  *  is  — 
2'         8  af'  gf       4  - 


44,  The  Radical  Sign,  ^,  is  also  used  to  indicate 
the  square  root  of  a  quantity.  When  any  other  than  the 
square  root  is  to  be  designated  by  this,  a  small  figure  speci- 
fying the  root  is  placed  in  the  sign.  Thus  V^  signifies 
the  3d,  or  cube  root  of  5,  and  is  the  same  as  5^.  \/Mal]^ 
indicates  the  5th  root  of  34a^  and  is  the  same  as  (34a^^)^. 

Scholium. — Read  \/5ac  ("the  square  root  of  5ac,"  not  "radical 
5ac'').  The  latter  expression  is  generic,  and  applies  as  well  to 
\/5ac,  or  \/5nc.     Besides  it  is  inelegant. 

Let  the  pupil  read  the  following  examples  and  give  the 
signification  of  each. 


18  FUNDAMENTAL    RULES. 


EXAMPLES. 

Ex.  1.  25a~^k  Bead,  *'25,  a  exponent  —2,  h  exponent 
|."  It  means  25  multiplied  by  -g,  multiplied  by  the  square 
of  the  cube  root  of  h.     (See  articles  109,  110.) 

2.  x^~\    Read,  "ic,  exponent 1."     Since 1  is 

^  n  n 

— ; — ,  x^~   is  the  same  as  x~^y  and  hence  means  that  x  is 

to  be  raised  to  a  power  indicated  by  m—n,  and  the  nth.  root 
of  this  power  extracted. 

3.  Read  and  explain  2a^b~^.    V27f*^y  « . 

SYMBOLS    OF    RELATION. 

45,  The  Sign  of  Geometrical  Ratio  is  two  dots 
in  the  form  of  a  colon,  :  .  Thus  a  :  h,  is  read  "a  is  to  h," 
or,  "the  ratio  of  a  to  &."     It  means  the  same  n,^  a-—h. 

46,  The  Sign  of  Arithmetical  Ratio  is  two  dots 
placed  horizontally,  ••  ,  Thus  a  -•  h  is  read,  "the  Arith- 
metical ratio  of  «  to  5  and  is  equiyalent  to  a—b. 

47,  The  Sign  of  Equality  is  two  parallel  horizontal 
lines,  = .  Thus,  2cx  ■=  xy,  is  read,  "  2cx  equals  xy.^^  6ac 
—2by  =  dx^  is  read,  "  oac  minus  2by  equals  3a:l" 

Four  dots  in  the  form  of  a  double  colon,  : : ,  is  the  sign 
of  equality  between  ratios.  Thus,  a:b::c:d.  read,  "a  is 
to  5  as  c  is  to  rZ,"  means  that  the  ratio  of  a  to  5  equals  the 
ratio  of  c  to  cl,  and  may  just  as  well  be  written  a  :  b  :=:  c  :d, 

a       c 
OY  ~=z  ~    all  of  which  expressions  mean  exactly  the  same 

thing. 

48,  The  Sign  of  Inequality  is  a  character  somewhat 
like    a    capital    V    placed  on  its  side,    <,   the  opening 


NOTATION.  19 

being  towards  the  greater  quantity.  Thus  ay  b  is 
read,  "a  greater  than  ^."  m  <i  n  is  read,  "m  is  less 
than  w." 

49.  The    Sign    of    Variation    is    somewhat    like    a 

figure  8  open  at  one  end  and  placed  horizontally.     Thus, 

c  c 

a  X  -,  is  read,  "  a  varies  as  ->." 
d  a 

SYMBOLS    OP    AGGREGATION. 

50.  A  Vinculum  is  a  horizontal  line  placed  over  several 
terms,  and  indicates  that  they  are  to  be  taken  together. 
The  parenthesis,  (  ),  the  brackets,  [  ],  and  the  brace,  i   I 
have  the  same  signification. 

Illustration.  a-\-l)y.cd—e  means  that  (a +  5)  is  to  be  multiplied 
by  {cd—e).  (a  +  &)  x  {cd—e)  also  means  the  product  of  {a  +  l)  and 
{cd—e).  Brackets  and  braces  are  used  when  one  parenthesis  would 
fall  within  another.  Thus,  \z-it[a-k-{])-\-c)x\y\  w,  signifies  that  the 
product  of  (&  +  6")  multiplied  by  x,  is  to  be  added  to  a,  and  this 
sum  multiplied  by  y\  to  this  product  z  is  to  be  added  and  the  sum 
multiplied  by  u. 

ol,  A  vertical  line  after  a  column  of  quantities,  each 
having  its  own  sign,  signifies  that  the  aggregate  of  the  col- 
umn is  to  be  taken  as  one  quantity.  Thus  -}-a  x  is  the 
same  as  {a—b-^c)x.  —h 

SYMBOLS    OP    CONTINUATION. 

32.  A  series  of  dots, ,  or  of  short  dashes, 

,  written  after  a  series  of  expressions,  signifies  ''&c." 

Thus  a  :  ar  :  ar^ :  ar^ ar^  means  that  the  series  is 

to  be  extended  from  ai^  to  ar^,  whatever  may  be  the  value 
of  n 


20  FUNDAMENTAL    RULES. 


SYMBOLS    OF    DEDUCTION. 

53,  Three  dots,  two  being  placed  horizontally  and  the 
third  above  and  between,  .  • . ,  signify  therefore,  or  some 
analogous  expression.  If  the  third  dot  is  below  the  first 
two,  •.-,  the  symbol  is  read  ^* since,"  *^  because,"  or  by 
some  equivalent  expression. 

POSITIVE    AND    NEGATIVE    QUANTITIES. 

54,  Positive  and  Negative  are  terms  primarily  ap- 
plied to  concrete  quantities  which  are,  by  the  conditions 
of  a  problem,  opposed  in  character. 

Illustration. — In  estimating  the  value  of  a  person's  estate,  bis 
property  may  be  called  positive,  and  his  debts  negative.  Distance 
up  may  be  called  positive,  and  distance  down,  negative.  Time 
before  a  given  period  may  be  called  positive,  and  after,  negative. 
Degrees  above  0  on  the  thermometer  scale  are  called  positive,  and 
below,  negative. 

55,  The  signs  +  and  —  are  used  to  indicate  the  char- 
acter of  quantities  as  positive  or  negative,  as  well  as  for  the 
purpose  of  indicating  addition  and  subtraction.  (See 
article  57.) 

56,  In  problems  in  which  the  distinction  of  positive 
and  negative  is  made,  each  quantity  is  to  be  considered 
as  having  a  sigti  of  character,  expressed  or  understood, 
besides  the  plus  or  minus  sign,  which  indicates  whether 
it  is  to  be.  added  or  subtracted.  The  positive  sign  need 
not  be  written  to  indicate  character,  as  it  is  customary  to 
consider  quantities  whose  character  is  not  specified  us 
positive. 

Illustration  I. — In  the  expression  aJ)-\-m—cx,  let  the  problem  out 
of  which  it  arose  be  such,  tliat  a,  m,  and  x,  tend  to  a  positive  result, 
and  b  and  c  to  an  opposite,  or  a  negative  result.     Giving   these 


NOTATIOT^.  21 

quantities  their  signs  of  character,  we  have,  (  +  a)  x  (~h)  -f  (+?//) 
—  (-c)  X  i  +  x,)  which  may  be  read,  ''positive  a  multiplied  by 
negative  b,  plus  positive  m,  minus  negative  c  multiplied  by  posi- 
tive a;."  Suppressing  the  positive  sign,  this  may  be  written, 
a  {-b)  +  771  —  (-c)  x^  by  also  omitting  the  unnecessary  sign  of 
multiplication. 

Illustration  2. — As  this  subject  is  one  of  fundamental  importance, 
let  careful  attention  be  given  to  some  further  illustrations.  We  are 
to  distinguish  between  discussi(ms  of  the  relations  between  mere 
abstract  quantities,  and  problems  in  which  the  quantities  have 
some  concrete  signification.  Thus,  if  it  is  desired  to  ascertain 
the  sum  or  difference  of  468,  or  //^,  and  327,  or  7i,  as  mere  num- 
bers, the  question  is  one  conceniiug  the  relation  of  abstract 
numbers,  or  quantities.  No  other  idea  is  attached  to  the  expres- 
sions than  that  each  represents  a  certain  number  of  units.  But, 
if  we  ask  how  far  a  man  is  from  his  starting  point,  who  has  gone, 
first,  468,  or  tt?,  miles  directly  east,  and  then  327,  or  n,  miles 
directly  west ;  or  if  we  ask  what  is  the  difference  in  time  between 
468,  or  m,  years  B.  C,  and  327,  or  ??,  years  A.  D.,  the  numbers  468, 
or  7//,  and  327,  or  ??,  take  on,  besides  their  primary  signification  as 
quantities,  the  additional  thought  of  opposition  in  direction.  They 
therefore  become,  in  this  sense,  concrete. 

Again,  a  company  of  5  boys  are  trying  to  move  a  wagon. 
Three  of  the  boys  can  pull  75,  85,  and  100  pounds  each  ;  and 
they  exert  their  strength  to  move  the  wagon  east.  The  other  two 
boys  can  pull  90  and  110  pounds  each;  and  they  exert  their 
strength  to  move  the  wagon  west.  It  is  evident  that  the  75,  85, 
and  100  are  quantities  having  an  opposite  tendency  from  90  and 
110.  Again,  suppose  a  party  rowing  a  boat  up  a  river.  Their 
united  strength  would  propel  the  boat  8  miles  per  hour  if  there 
were  no  current;  but  the  force  of  the  current  is  sufficient  to  carry 
the  boat  2  miles  per  hour.  Which  way  will  the  boat  move,  and 
how  fast  ?  The  8  and  2  are  quantities  of  opposite  character  in  their 
relation  to  the  problem.  Once  more,  in  examining  into  a  man's 
business,  it  is  found  that  he  has  a  farm  worth  m  dollars,  personal 
property  worth  n  dollars,  and  accounts  due  him  worth  c  dollars. 
There  is  a  mortgage  on  his  farm  of  &  dollars,  and  he  owes  on  account 
a  dollars.  The  m^  «,  and  c  are  quantities  opposite  in  their  nature 
to  b  and  a.  This  opposition  in  character  is  indicated  by  calling  those 
qiuintities  which  contribute  to  one  result  positive^  and  those  which  con- 
tribute to  the  opposite  result  negative. 


32  FUNDAMENTAL    RULES. 

^7.  Purely  abstract  quantities  have,  properly,  no  dis- 
tinction as  positive  and  negative ;  but,  since  in  such  prob- 
lems the  plus  or  additive,  and  the  minus  or  sub  tractive 
terms  stand  in  tlie  same  relation  to  each  other  as  positive 
and  negative  quantities,  it  is  customary  to  call  them  such. 

Illustration. — In  the  expression,  5ac—3cfZ-|-8a'y—2a^,  though  the 
quantities,  «,  c,  d,  x  and  y  be  merely  abstract,  and  have  no  proper 
signs  of  character  of  their  own,  the  terms  do  stand  in  the  same 
relation  to  each  other  and  to  the  result,  as  do  positive  and  negative 
quantities.  Thus,  5ac  and  ^xy  tend,  as  we  may  say,  to  increase  the 
result;  while  —Zed,  and  —  2atf?  tend  to  diminish  it.  Therefore  the 
former  may  be  called  positive  terms,  and  the  latter  negative. 

*5S,  Scholium. — Less  than  zero.  Negative  quantities  are  fre- 
quently spoken  of  as  "less  than  zero."  Though  this  language  is 
not  philosophically  porrect,  it  is  in  such  common  use,  and  the  thing 
signified  is  so  sharply  defined  and  easily  comprehended,  that  it  may 
possibly  be  allowed  as  a  conventionalism.  To  illustrate  its  mean- 
ing, suppose  in  speaking  of  a  man's  pecuniary  affairs  it  is  said  that 
he  is  worth  "  less  than  nothing ;"  it  is  simply  meant  that  his  debts 
exceed  his  assets.  If  this  excess  were  $1000,  it  might  be  called 
negative  $1000,  or  -$1000.  So,  again,  if  a  man  were  attempting  to 
row  a  boat  up  a  stream,  but  with  all  his  effort  the  current  bore  him 
down,  his  progress  might  be  said  to  be  less  than  nothing,  or  nega- 
tive. In  short,  in  any  case  where  quantities  are  reckoned  both 
ways  from  zero,  if  we  call  those  reckoned  one  way  greater  than 
zero,  or  positive,  we  may  call  those  reckoned  the  other  way  "  less 
than  zero,"  or  negative, 

SO,  The  value  of  a  Negative  Quantity  is  conceived  to 
increase  as  its  numerical  value  decreases. 

Illustration. — Thus  -3  >  -5,  as  a  man  who  is  in  debt  $3,  is  bet- 
ter off  than  one  who  is  in  debt  $5,  other  things  being  equal.  If  a 
man  is  striving  to  row  up  stream,  and  at  first  is  borne  down  5  miles 
an  hour,  but  by  practice  comes  to  row  so  well  as  only  to  be  borne 
down  3  miles  an  hour,  he  is  evidently  gaining;  i.  e.,  -3  is  an  in- 
crease upon  -5.  Finally,  consider  the  thermometer  scale.  If  the 
mercury  stands  at  20°  below  0,  (marked  -20°)  at  one  hour,  and  at 
-10'  the  next  hour,  the  temperature  is  increasing;  and,  if  it  in- 
crease sufllciently  will  become  0,  ja^dng  which  it  will  reach  +1°, 


NOTATION.  23 

+  2°,  etc.  In  this  illustration,  the  quantity  passes  from  negative  to 
positive  by  passing  through  0.  This  is  assumed  as  a  fundamental 
truth  of  the  doctrine  of  positive  and  negative  quantities,  viz. :  That 

A   QUANTITY    IN    PASSING    TUROUGH   0    MAY    CHANGE    ITS   SIGN. 

[It  appears  in  geometry,  that  a  quantity  may  also  change  its  sign 
in  passing  through  infinity.  Thus  the  tangent  of  an  arc  less  than 
90°  is  positive;  but  if  the  arc  continually  increases,  the  tangent 
becomes  infinity  at  90°,  passing  which  it  becomes  negative.] 


NAMES  OF  DIFFERENT  FORMS  OF  EXPRESSION. 

00.  A  Polynomial  is  an  expression  composed  of  two 
or  more  parts  connected  by  the  signs  plus  and  minus,  each 
of  which  parts  is  called  a  term. 

61.  A  Monomial  is  an  expression   consisting   of  one 
term. 
A  Binomial  is  a  polynomial  having  two  terms. 
A  Trinomial  is  a  polynomial  having  three  terms. 

Illustration.  5a''h—cd»+x—4.{a  +  l)  is  a  polynomial  of  4  terms. 
The  first  three  are  monomial  terms,  and  the  last  is  a  binomial  term. 
5a4i-~ef  and  x^  +  y'  are  examples  of  binomials.  'Za'^x^y—  125<Z-"'  + 12 
is  a  trinomial. 

02.  A  CoeflBcient  of  a  term  is  that  factor  which  is 
considered  as  denoting  the  number  of  times  the  remainder 
of  the  term  is  taken. 

The  numerical  factor,  or  the  product  of  the  known  fac- 
tors in  a  term  is  most  commonly  called  the  coefficient, 
though  any  factor,  or  the  product  of  any  number  of  factors 
in  a  term  may  be  considered  as  coefficient  to  the  other 
part  of  the  term. 

Illustration. — In  the  term  6a,  6  is  the  coefl5cient  of  a.  In  ax,  a 
may  be  called  the  coeflBcient  of  a;,  or  1  may  be  called  the  coefficient 
of  ax.  In  Qaxy,  6  is  the  coefficient  of  axy,  Q^a  of  xy,  and  (Sax  of  y. 
In  5  (oft— c),  5  is  the  coefficient  of  (ab—c) ;  and  in  (2a^—cd)xy,  (2a' 
—ed)  is  the  coefficient  o^xy. 


24  FUNDAMENTAL    RULES. 

63,  Similar  Terms  are  such  as  consist  of  the  same  let- 
ters affected  with  the  same  exponents. 

Illustration.  5a6,  13«5,  and  db  are  similar,  —l^x^y's^  5a;'2^,  and 
—x^y^  are  also  similar  to  each  other.  4a&,  5a5^  and  —  2,db^  are  dis- 
similar, as  are  8aaj,  —^hx^^  4ca!^,  and  ^xy. 


EXERCISES    IN    NOTATION. 

Ex.  1.  Write  in  mathematical  symbols,  5  times  the  square 
root  of  a,  added  to  the  cube  of  the  sum  of  a  and  h. 

Result,  («  +  ^)3  +  5  v^a,  or  {a  +  iy-^ba^. 

How  many  terms  in  this  result  ?  What  kind  of  a  term  is 
the  first  one  ? 

2.  Write  the  second  power  of  a,  plus  3  times  the  product 
of  c  square  multiplied  by  h,  diminished  by  m  times  the 
cube  root  of  the  binomial,  the  square  root  of  a  minus  the 
cube  of  I. 

Result,  a^-\-2,c^b-m{a^^—b^)^,  or  a?-\-?>(^h—m.  ^aV^-l^. 

3.  Write  three  times  a  into  h,  plus  the  binomial  a  minus 
h,  divided  by  the  sum  of  a  square  and  h  cube. 

4.  Write  the  fraction,  the  product  of  the  sum  of  a  and  I 
into  the  sum  of  x  and  y,  divided  by  the  square  root  of  a 
diminished  by  the  cube  root  of  h. 

va—\h 

6.  Write  the  fraction,  a  fifth  power  diminished  by  3 
times  a  square  h  cube,  divided  by  the  square  root  of  the 
binomial  x  square  diminished  by  y  square. 

6.  Write  the  square  root  of  the  sum  of  x  and  y  equals  c 
minus  the  square  root  of  the  sum  of  x  and  h. 

7.  Write  the  fraction,  the  binomial  3  times  x  plus  1, 
divided  by  5  times  x,  minus  the  fraction  3  times  the  bino- 


NOTATION.  25 

mial  X  minus  1,  divided   by  the  binomial  3x  plus  2,   is 
greater  than  9  divided  by  11a:. 

8.  Write  the  square  root  of  the  fraction,  h  divided  by  a 
plus  Xy  plus  the  square  root  of  the  fraction  c  divided  by  a 
minus  x,  equals  the  4th  root  of  the  fraction,  4  times  the 
product  of  b  and  c,  divided  by  a  square  minus  x  square. 

9.  Write  a  exponent  |,  minus  b  exponent  —m.  Write  the 
result  in  three  different  forms. 

10.  Write,  a  exponent  —  ,  is  to  the  binomial  h  minus  x 

exponent  —  |,  as  5  times  c  square  plus  rf,  is  to  the  5th  root 

of  X  4th  power. 

*  1  _ 

Result,  an  :  -jt==-  : :  5^  +  (7  :  \^x*, 
W(b—xY 

What  binomials  are  there  in  the  last  result  ? 


EXERCISES    IN    READING    AND    EVALUATING 
EXPRESSIONS. 

Read  the  following  expressions,  and  find   the  value  of 
ejich,  when  a  =  6,  S  =  5,  r  =  4,  and  ^  =  1. 

1.  a'^  +  Ub—c-\-d.  Result,  3G 4-60—44-1,  or  93. 

2.  2«3_3fl2^  +  c3.  Result,  —44. 

3.  3(a2-//J)-«(cf-h^).  Result,  15. 

4.  Between   the  expression   -^ — -.  and    wba^—{(^-{-^b), 

which  of  the  signs,  =,  >,  or  <,  is  correct  ? 

Read  and  evaluate  the  following,  calling  «  =  IC,  &  =  10, 
c  =z  Q,  m  r=  A,  X  z=z  5,  y  =  1. 


5.  {b—x){Va  +  b)-\-y/{a—b)(x-\-y),  Result,  76. 

2 


26  PtJNDAMEKTAL    RULES. 

6.  VcJa  +  b)  —  ^/¥{a—b).  T^esw//,  6.49,  nearly. 

7.  50«a;'^4-4a^-100[<x-(2aJ  +  |)l-       Result,   -112. 

8.  3a-*  +  4  i^-^^f-,-  ^  +  y\  B.SUIU  41. 
Suggestion. — Any  power  of  1  is  1. 

9.  ^x-^^^  |!     «--^^.  Resulty  2451. 


10.  Find  the  value  of  aV^^—^ci-]-xVx^-\-^a,  when 
a;  =  5,  and  «  =  8. 

IL  Find  the  value  of  a  +  h  ^{x-^y)  —  (a  —  'b)  ^{x—y), 
when  a  zz=  10,  5  =  8,  a;  ==  12,  and  ?/  =  4. 

Suggestion.  V(^  +  2/)  is  equivalent  to  ^^'x->^y,  the  parenthesis 
and  the  vinculum  having  the  same  signification. 

12.  If  a  =  2,  ^  =  3,  2;  =  6,  and  y  =  5,  show  that 
V\{a-^b)y\-}-\^\{a  +  x){y-2a)\-hv'\{y-bya\=9. 

13.  With  the  same  values  show  that  {ay)^  (^a;  +  «^  +  3)~« 
_  \b(x-y)-^-[(axY-U2]  \  20ay  _ 

14.  Find  the  value  of  ^/W^^  —  {lO-]-n)^  if  w  =  6. 

15.  Find  the  value  of  (5771^  +  5^/^)^+^/^+0;^,  if 
771  :=  4,  and  a;  =  9. 

Test  Questions. — Wliat  are  the  chief  points  of  difference  between 
the  Arabic  or  Decimal  notation,  and  the  Literal  or  Algebraic  ? 
What  is  meant  by  the  terms  positive  and  negative  as  applied  to 
quantity  ?  What  is  the  meaning  of  the  negative  sign  when  prefixed 
to  an  exponent  ?     Read 


oc  ,  or  =,  or  <,  or  >  y  wa?— y"  :  4m^. 


NOTATION. 


37 


SYNOPSIS    FOR    REVIEW. 


DEFINITION. 


SYMBOLS   OF 
QUANTITY. 


Arabic. 


Literal. 


See  Arithmetic. 


Advantages. 
Examples, 


Ist  Law. 


2nd  Law. 


SYMBOLS   OF       ,    „„„^ 
OPERATION  AND    \    Exponent. 
DEFINITIONS.  *^ 


Infinity.         Meaning  of. 

+  ,  —    vT,    X,   - 
Definitions  of 
Power, 
Root, 


j  1.  More  general. 

i  2.  Can  (race  quantities. 

(  Known  quantities. 
Unknown  quantitie>i. 
Accents.,  t^ubccriptt:,  and 

Greek  Udtefrs. 
Difference  beticeen  Alge- 
l>raanct  Arithmetic. 

Letters  in  connection. 
Two  points  of  difference. 
Figures  with  letters. 


b)a{c, 


a\b 


V. 


SYMBOLS  OF 

RELATION     AND 

DEFINITIONS. 


SYMBOLS   OF 
AGGREGATION. 


Integral  Exponent, 
Fractional  Exponent, 
Negative  Exponent, 
Radical  Sign. 

:,  ••,  =,  ::,  >  <,  a. 
Geometrical  Ratio. 
Arithmetical  Ratio. 
Equality. 
Inequality. 
Variation. 

r   — .  ().  [].  iK  I. 

J     Vinculum. 

l^    Vertical  Line. 


How  rt-fu\.    Examples. 


SYMBOLS  OF  CONTINUATION. 
SYMBOLS  OF  DEDUCTION. 


POSITIVE  AND 

NEGATIVE 
QUANTITIES. 


FORMS  OF 
EXPRESSION. 


Definition. 

Two  signs  of  every  quantity. 

Plus  and  minus  terms  become  positive  and  negative. 

(  Meaning. 
"  Less  than  zero."  -<  How  negatives  increase. 

'  How  a  quantity  changes  sign 


Polynomial. 

Monomial. 

Binomial. 

Trinomial. 

CoeflRcient. 

Similar  Terms. 


Term. 


Illustrations. 


EXERCISES   IN. 

EXERCISES   IN    READING   AND   EVALUATING   EXPRESSIONS. 


2.8 


FUNDAMENTAL    RULES. 


64.  Addition  is  the  process  of  combining  several  quan- 
tities, so  that  the  result  shall  express  the  aggregate  value 
in  the  fewest  terms  consistent  with  the  notation. 

60,  The  Sum  or  Amount  is  the  aggregate  value  of 
several  quantities,  expressed  in  the  fewest  terms  consistent 
with  the  notation. 

Illustration. — To  add  346,  234,  and  15,  is  to  find  an  expression 
for  their  aggregate  value  in  the  fewest  terms  consistent  with  the 
decimal  notation.  The  sum  or  amount  is  595,  because  it  is  such 
simplest  expression  for  the  aggregate.  In  like  manner  the  sum  of 
4ac  +  51}  +  2x,  idac  +  2h  +  Sx,  and  121)  +  9x,  is  17«c+19a;+19&,  be- 
cause it  is  the  simplest  expression  for  the  aggregate  value  consistent 
with  the  literal  notation. 

If  the  pupil  is  acquainted  with  other  scales  of  notation  he  knows 
that  with  radix  100,  595  would  be  represented  by  2  figures,  since 
all  numbers  less  than  100  would  be  represented  by  one  figure. 


00.  Prop.  1.  —  By  Addition  similar  terms  are 
united  into  one. 

Demonstration. — Let  it  be  required  to  add  4«c,  5ac,  —2ac,  and 
— 3ac.  Now  4«c  is  4  times  ac,  and  5«c  is  5  times  the  same  quantity 
{ac).  But  4  times  and  5  times  the  same  quantity  make  9  times 
that  quantity.  Hence,  4«c  added  to  5«c  make  ^ac.  To  add  —2ac 
to  9ac  we  have  to  consider  that  the  negative  quantity,  — 2ac.  is  so 
opiJosed  in  its  character  to  the  positive,  9«c,  as  to  tend  to  destroy 
it  when  combined  (added)  with  it.  (As  if  'doc  were  property,  and 
—2ac  debts.)  Therefore,  —2a^  destroys  2  of  the  9  times  axi,  and 
gives,  when  added  to  it,  llac.     In  like  manner  — 3«c  added  to  lac. 


ADDITION.  29 

gives  4^c.  Thus  the  four  similar  terms,  iac,  5ac,  —2ac,  and  —3ac, 
have  been  combined  (added)  into  one  term,  Aac ;  and  it  is  evident 
that  any  other  group  of  similar  terms  can  be  treated  in  the  same 
manner     q.  e.  d. 

EXAMPLES. 
Ex.  1.  Add  13m%,  —lOmhi,  —Cmhi,  omhi,  and  —im^n. 

Model  Solution.— Adding    together    13m'^/i    and    —10m'/?,   the 

—  lOw'w  destroys  10  of  the  13  times  m^n  and  gives  3m'w.  Adding 
3m'/i  and   —Qm^n,  the  'dm^n  destroys  3  of  the  —Q/u^n  and  gives 

—  dm'?}.  —Sm'^n  added  to  5m^n  destroys  3  of  the  5  times  m'/i 
and  gives  2?n'^n.  2m^n  added  to  — 4m'n  destroys  2  of  the  — 47?t'n 
and  gives  — 2m'w.  Hence  the  sum  of  13m"^r?,  —lOm^n,  —  6m'w, 
57n'n,  and  —4'm?n  is   —2m'^n. 

2.  Add  ISax^,  —5ax^,   —lOax^,  4:ax^,  and  —6ax^,  ex- 
plaining as  above.  Result,  ax^. 

3.  Add   —bSx?,   —)lc^x\  %ch^,  Sch^,  and  —ic*x%  ex- 
plaining as  before.  Result,  0. 

4.  Add  3fl.r,  Qax,  —ax,  2ax,  —7 ax,  and  6ax, 

5.  Add  2%^  —iibf,  —hyS  Sby\  3b f,  and  —2by\ 

6.  Add  5ax-,  —2ax^,  dax^,  —9aa^,  and  ax^. 

Sum,   —2ax\ 

7.  Add  6  J,  —6x^,  —  lOar^,  Sx^,  and  lla;i 

8.  Add  —6a^  2a%  —oa^,  \a?,  — 3«2,  and  ^2. 

9.  Add  —2a^/x,  a^/x,  —SaVx,  la^fx,  and  —^a^/x. 

Sum,  —a^/x. 

10.  Add  —  2am^,  4a\/wi,  3am^,  and  —  aa/w. 

/S^wm,  4«'v/wi. 

11.  AddlOflW,  -^a^^/x,  -2aW,  and  4a*^i. 

12.  Add  \\am,  %\am,  —3am,  and  am.  Sum,  2am, 


30  FUNDAMENTAL    RULES. 

13.  Add  27^^  —a\  — 28^2,  and  — 4^^.       Sum.   — 6««. 

14.  Add  3?n2,  — fm^,  m^,  and  — |m2.  ^9i/w,  2-^m\ 
16.  Add  7a/5,  — 5a/^,  1%Vx,  and  — 3v^.  6^2^77?,  11 V^. 

16.  Add  9^,  }J,  — |J,  —85,  and  — 15.        .%m, .— ^. 

G7»  Cor.  1. — 7n  adding  similar  terms,  if  the  terms  are  all 
podtive,  the  sum  is  positive;  if  all  negative,  the  sum  is 
negative ;  if  some  are  positive  and  some  negative,  the  sum 
takes  the  sign  of  that  kind  (positive  or  negative)  which  is  the 
greater. 

Scholium. — The  operation  of  adding  positive  and  negative  quan- 
tities may  look  to  the  pupil  like  Subtractiou.  For  example,  we 
say  +5  and  -3  added  make  +2.  This  looks  like  Subtraction,  and 
in  one  view,  it  is  Subtraction.  But  why  call  it  Addition?  The 
reason  is,  because  it  is  s,imp\j  putting  the  quantities  together — aggre- 
gating them— not  finding  their  difference.  Thus,  if  one  boy  pulls 
on  his  sleigh  5  pounds  in  one  direction,  while  another  boy  pulls 
8  pounds  in  the  opposite  direction,  the  combined  (added)  effect  is 
3  pounds  in  the  direction  in  which  the  first  pulls.  If  we  call  the 
direction  in  which  the  first  pulls,  positive,  and  the  opposite  direction 
negative,  we  have  +5  and  -3  to  add.  This  gives,  as  illustrated,  +5i. 
Hence  we  see,  that  the  sum  of  +5  and  -3  is  +2. 

But  the  diff'erence  between  +5  and  —3  is  8,  as  appears  in  the 
following  illustration :  Suppose  one  boy  is  drawing  his  sleigh  for- 
ward while  another  is  holding  back  3  lbs.  If  it  takes  just  10  lbs. 
to  move  the  sleigh  itself,  the  first  boy  will  have  to  pull  13  lbs.  to 
get  it  on.  But  if  instead  of  holding  dacJc  3  lbs.,  the  second  boy 
pushes  5  lbs.,  the  first  boy  will  only  have  to  pull  5  lbs.  Thus  it 
appears,  that  the  difference  between  pushing  5  lbs.  (or  +  5)  and 
holding  back  3  lbs.  (—3)  is  8  lbs. 

In  like  manner  the  sum  of  $25  of  property  and  $15  of  debt,  that 
is  the  aggregate  value  when  they  are  combined,  is  $10.  +25  and 
-15  are  +10.  But  the  difference  between  having  $25  in  pocket,  and 
being  $15  in  debt,  is  $40.   The  difference  between  +  25  and  - 15  is  40. 

17.  A  thermometer  indicated  +28°  (28°  above  0),  it 
then  rose  10°,  then  fell  3°,  then  rose  2°,  and  again  fell  7°. 
What  was  the  sum  of  its  movements  ;  or,  how  did  it  stdJid 
at  last  ? 


ADDITION.  31 

Model  Solution. — Calling  upward  movement  +  and  downward 
-,  the  movements  were  +10,  -3,  +2,  and  -7,  the  sum  of  which  is 
+  2.  Hence  it  rose  2°.  As  it  originally  stood  at  28°  above  0,  and, 
in  the  whole,  rose  2\  it  stands  at  last  30°  above  0. 

18.  A  party  are  rowing  up  a  stream,  and  alternately  row 
and  rest.  During  3  periods  of  rowing  they  advance  Smn, 
2mnj  and  6mn  rods.  But  during  the  corresponding  periods 
of  resting,  they  float  down  6mn,  mn,  and  ^mn  rods.  What 
was  the  result ;  did  they,  on  the  whole,  ascend  or  descend, 
and  how  much?  In  other  words,  what  is  the  sum  of 
-r'Smiij  +2m/?,  -f-6mn,  — omn,  — mn,  and  — Amn? 

Arts.  They  ascended  mn  rods.     {-\-mn.) 

19.  A  man  has  a  farm  worth  llOOc^,  on  which  there  is  a 
mortgage  of  tlbcd ;  he  has  personal  property  worth  $8a/, 
and  accounts  due  him  of  %%cdy  but  owes  on  account  ^ocd, 
and  on  note  ^Icd.  What  is  the  sum  of  his  effects  ?  Or 
what  is  the  sum  of  +100cc?,  —Ibcd,  -\-Scd,  -{-2cd,  —bed, 
and  —Kcd?  Ans.  He  is  worth  $Sdcd.     {-}-H'3cd.) 

OS,  Cor.  2. — The  sum  of  two  quantities,  the  one  positive 
and  the  other  negative,  is  the  numerical  difference,  with  the 
sign  of  the  greater  prefixed. 

GO,  Cor.  3. — It  appears  that  addition  in  mathematics  does 
not  always  imply  increase.  Whether  a  quantity  is  increased  or 
diminished  by  adding  another  to  it  depends  upon  the  relative 
nature  of  the  two  quantities.  If  they  both  tend  to  the  same  end, 
the  result  is  an  increase  in  that  direction.  If  they  tend  to  oppo- 
site ends,  the  result  is  a  diminution  of  the  greater  by  the  less. 


70,  Prop.  2. — Dissimilar  terms  cannot  he  united 
into  one  by  addition,  but  the  operation  of  adding  is 
represented  by  writing  them  in  succession,  the  positive 
terms  being  preceded  by  the  +  sign  and  the  negative 
by  the  —  sign. 


32  .  FUNDAMENTAL    RULES. 

Demonstration. — Let  it  be  required  to  add  +4cy',  -\-Mb,  —2xt/, 
and  —mn.  Acy"^  is  4  times  cy^,  and  Sal  is  3  times  ab,  a  diflferent 
quantity  from  cy^ ;  the  sum  will,  therefore,  not  be  7  times,  nor,  so 
far  as  we  can  tell,  any  number  of  times,  cy'^  or"  ab,  or  any  other 
quantity,  and  we  can  only  represent  the  addition  thus :  4:Ci/^  +  Sab.  In 
like  manner,  to  add  to  this  sum  —2xy  we  can  only  represent  the 
addition,  as  4:cy^  +  Sab  +  (—  2xy).  But  since  2xy  is  negative,  it  tends 
to  destroy  the  positive  quantities  and  will  take  out  of  them  2xy. 
Hence  the  result  will  be  A:mf +  Sdb—2xy.  The  effect  of  —mn  will  be 
the  same  in  kind  as  that  of  —2xy^  and  hence  the  total  sum  wdll  be 
4:cy^  +  dab—2xy—mn.  As  a  similar  course  of  reasoning  can  be 
applied    to    any    case,    the    truth    of   the    proposition  appears. 

Q.  E.  D. 

Scholium. — In  such  an  expression  as  4:cy^  +  dab—2xy — mn,  the  — 
sign  before  the  um  does  not  signify  that  it  is  to  be  taken  from  the 
immediately  preceding  quantity ;  nor  is  this  the  signification  of 
any  of  the  signs.  But  the  quantities  having  the  —  sign  are  consid- 
ered as  operating  to  take  away  so  much  from  atiy  which  may  have 
the  +  sign,  and  vice  versa. 


EXAMPLES. 
Ex.  1.  Add  together  bax,  —lOcy,  Sb,  and  —n. 

IVIodei  Solution.  5ax  and  lOcy  being  dissimilar  will  not  unite 
into  one  term,  since  one  is  5  times  ax,  and  the  other  is  10  times  cy, 
a  different  quantity ;  therefore  I  can  only  rej^resent  the  addition,  as 
bax-\-  {—\Qcy).  But  the  lOcy  being  negative  tends  to  offset 
positive  quantities,  and  will  take  out  of  such  its  own  value. 
Hence  bax  +  (—  lOcy)  is  ^ax  —  lOcy.  To  this  adding  8&, 
which  is  positive  and  hence  will  go  to  increase  the  result,  I 
have  ^ax  —  lOcy  +  8&.  Finally,  as  n  is  negative  it  diminishes  the 
result  by  its  numerical  value,  and  I  have  for  the  sum  ^ax—lQcy 
+  Sb—n. 

2.  Add  together  4«m,  —  2c%,  —8x,  and  5Jn,  explaining 
as  above. 

3.  Add  —  2c2m*,  4:cm,  —  6cW,  and  lOc^m,  explaining 
as  before. 


ADDITION.  88 

4.  Add  la'^,  —  32ir"^,  Qmn,  and  —6a^,  explaining  as 
before,  and  find  the  numerical  value  of  the  result  if  a  =  3, 
J  =  18,  a;  =  8,  m  =  2,  and  n  =  5.  Result,  21. 

7  J.  Cor. — Adding  a  ner/aiive  quantity  is  the  same  as  sub- 
tracting a  numerically  equal  positive  quantity ;  that  is, 
m  +  (— n)  is  m  — n. 


7*^.  Prob. — To  add  Polynomials. 

Rule. — /.  Write  the  polijnoniials  so  that  similar 
terins  shall  fall  in  the  same  coluimi. 

II.  Combine  each  column  into  one  term,  and  write 
the  result  underneath  luith  its  own  sign. 

TJie  polynomial  thus  found  is  the  sum  sought. 

Demonstration. — As  the  object  is  to  combine  the  quantities  into 
the  fewest  terms,  it  is  a  matter  of  convenience  to  write  similar  terms 
in  the  same  column,  as  such,  and  only  such,  can  be  united  into  one. 
(66,  "70.)  Now,  since  in  polynomials  the  plus  and  minus  terms 
Ktand  in  the  same  relation  to  each  other  as  positive  and  negative 
quantities  (57),  they  may  be  considered  as  such,  and  united  by  67. 
The  partial  sums  will  then  be  dissimilar  terms  and  will  be  added 
by  connecting  them  with  their  own  signs  (70).     Q.  e.  d. 

EXAMPLES. 

Ex.1.  Add  together  16ac—'2m-{-xy,  'dm  —  5xy—d—2aCf 
'-Sxy—4:ac—0m,  and  2mn—3ac-\-Sxy. 

Model  Solution. — Writing  the  first  polynomial  as  it  stands,  I 

arrange  the  others  so  that 
16ac  —  2m  +    xy  similar  terms  shall    stand 

—  2ac  +  3m  —  6xy  —  d  in  the    same    column,   for 

—  4ac  —  6m  —  3xy  convenience      in      uniting 

—  5^^; +  ^xy  +  2mn     ti^^.,^       rpj^^j.^    ^^^^^    ^^^ 

lac,—  5m,  +     '^'y?—  d,-^  2mn     term   similar   to    ^%mn   I 
lac  —  5m  4-    xy  —  d  -\-  2mn     bring  it  down,  and  in  like 


34  FUNDAMENTAL    RULB8. 

manner  — d.  8a^  and  —3xi/  are  5xt/.  5x1/  and  —5xy  are  0.  0  and 
ity,  or  simply  -{-xy,  is  the  sum  of  the  similar  terms  in  xy.  Writing 
this  result  I  pass  to  the  next  column.     —Qm  and   +3m  are  —Zm. 

—  dm  and  —2m  are  —5m  which  being  the  sum  of  the  similar  terras 
in  m,  is  written  down  In  like  manner  the  sura  of  the  terms  in  ac 
is  lac.  The  partial  sums  are,  therefore,  7ac,  —5m,  +xi/,  —d,  and 
+  2wn.  But  these  being  dissimilar  terms  are  added  by  connecting 
them  with  their  own  signs  (70) ;  whence,  the  sum  of  the  several 
polynomials  is  '7ac—5m-\-xy—d  +  2mn. 

In  like  manner  solve  and  explain  the  following  : 

2.  Add  Qx  +  5at/,  ^3x-\-2ay,  x—6ay,  2x-{-ay. 

Sum,  Qx-\-2(ty, 

3.  Add  day— 7,  —ay*-\-8,  2ay—9,  —Say— 11,  and  lOay 
—13.  Sum,  Hay— 32. 

4.  Add    —Sab  +  lx,  dab—lOx,  3ab—6x,  —ab-\-9x,  and 
2aJ)-\-^x,  Sum,  4:ab-^4:X. 

5.  Add  —6a^-^2I),   —3d  +  2a%  -ba^-Sh,  4:a^—2b,  and 
95— 3a2.  Sum,   —8a^—2b. 

6.  Add  SaW  —  7a¥  +  ^axy,    —  7aW  —  2a¥  —  axy,  a¥ 
—laxy  +  8am,  —l{)a¥-[.a^^+3axy,  and  —5a^b^-\-\8a¥. 

Sum,  0. 

7.  Add   ldax^—Uy^-\-3a(^—mn^,   4.ax^-{-15y^—3a%  ^y"^ 

—  17ax^-^2ac^-j-2m^n-{-3mn^,  and  lOy^—a^c. 

Sum,  15y^-^6ac^-]-2mn^—4:ah-^2m^n. 

8.  Find  the   sum  of  2a^-\-4:l^x—c^x%   2(^x^-{-^a^—6b% 
SLnd21^x—4:C^x^-]-2aK  Sum,  Sa^-Sc^a^. 

9.  Find  the   sum  of   8a^x^ — 3xy,  5ax—5xy,    9xy—5ax, 
2a^x^  +  xy,  siiad  6ax—3xy.  Sum,  lOa^x^-^bax—xy. 

10.  Find  the  sum  of  2bx—12,  3x^—2bx,  bx^—3Vx,  Wx 
+  12,  x^-\-3,  and  ox^—lVx,  Sum,  Ux^-Wx  +  3. 

11.  What  is  the  sum  of  20a^c^x-{-16ah—15a^(^x—23ah  ? 

Ans.,  5a^(^x—Sah. 


ADDITION.  35 

12.  What  is  the  sum  of  16x?/  —  Um  +  17xy  -f-  Wi7n? 

Ans.y  33xy  +  Sdhm. 

13.  What  is  the  sum  of  5c*  —  4dan  +  10c*  f  Uan  ? 

Ans.,  15c*  —  36an. 

14.  Wliat  is  the  sum  of  lOx^y^  —  176^-  +  l^x^y^  +  bsk 

—  4a:V^?  Ans.,  ^Ix^y^  —  nsk. 

Scholium  I. — In  practice,  the  expert  will  not  take  the  trouble  to 
arrange  the  polynomials,  but  will  simply  select  and  combine  the 
similar  terms,  writing  each  result  in  the  total  sum,  at  once.  Thus, 
in  solving  Ex.  7,  when  the  object  is  simply  to  find  the  sum,  and  not, 
as  above,  to  explain  the  jjrocess,  we  proceed  as  follows :  Noting  the 
13aa;^,  we  cast  the  eye  along  till  we  find  the  similar  term,  +4aa!^, 
and  say  "4-17aa;'';"    again,   casting  the   eye   along  till   we  find 

—  17aa;',  we  say  "0."  Therefore  nothing  is  written  in  the  sum  for 
these  terms,  as  they  mutually  destroy  each  other.  Again,  looking 
to  — 14y^,  and  then  on  to  I5y'\  we  say  "y'',"  and,  passing  on  to 
+  4y',  say  "oy'V'  ^^^  again  passing  on  to  lOy^  and  say  "  15y  ." 
This  being  the  sum  of  the  terms  in  y^,  it  is  then  written  in  the 
answer.  In  the  same  manner  the  work  is  carried  on  to  completion ; 
/.  e.,  only  naming  results^  and  writing  them  in  the  total  sum. 

In  this  manner,  write  the  answers  in  the  following  ex- 
amples : 

15.  Add  'dx  —  6y  -\-  4c,  2x  —  2y  —  3c,  and  —x-\-dy-i-c. 

Stim,  ix  —  4y  +  '2c. 

16.  Add    lla-j-  i3x  —  7d,    4:a  —  10x  —  2d,    and    —9a 

—  z  -\-  3d.  Sum,  6a  -{-  2x  —  iSd. 

17.  Add  dax  —  4tby  +  2?nn  —  16,  2by  —  57nn  +  11, 
'd7n7i  —  2ax  +  5,     and     —  by  -\-  m7i  —  ax. 

Su77i,  rriTi  —  Sdy. 

18.  Add  6a7nx  —  3b  +  ^cxy  —  2ax^,  ib  —  3cxy  +  bax^ 
llcxy  —xy  —  SaTUx  —  3ax^,  and  bxy  —  6  —  b. 

SuTTif  12cxy  —  2amx  -f  ^xy  —  0, 


36  FUNDAMENTAL    RULES. 

19.  Add  '^xy  —  2x^,  3x^  +  xy,  x^  +  xy,  and  4:X^  —  dyx. 

20.  Add  2ax  —  30,  Sa^  —  2ax,  5x^  —  3x^,  and  SV^+IO. 

Sum,  8a;2  —  20. 

21.  Add  Sd'x^  —  3«a;,  7 ax  —  6xy,  9xy  —  bax,  and  2d\x^ 

22.  Add  ^ax^-\-bVx,  —2axi^6xK  3aa^—10x^y  —Hax^ 
4-  'SVx,  and  «a;2  -f  11  Va;. 

23.  Add     6a;y  —  12a;2,    —  4ii;2  +  Sxy,    ^x^  —  2xy,    and 

—  3xy  +  4a;2. 

24.  Add  4:ax  —  130  +  dx^,  bx"^  +  dax  +  9^2,  7:cy  —  Wx 
+  90,  and  V^  +  40  -  6a:2. 

25.  Add  3r?-2  +  Uc  —  e^  +  10,  -  5«-2  +  6bc-\-2e'  -  15, 
and  _  4a-2  —  9^c  —  lOe^  +  21. 

26.  Add  la—bif,  ^^/x  +  2a,  by^—^/x,  and  ^^a-^-l^/x 
together.  Sum,  iWx. 

27.  Add  4:m7i-^3ab—4:C,  3x—4:ab -\- 2mn,  and  Sm^  —  Ap 
together.  Sum,  6mn  —  ad  —  4:C  i-  'Sx  -{-  3m^  —  4^. 

28.  Find  the  sum  of    3d^  +  2ab  +  4:b%    ba^  -  Sab  +  ^, 

—  ^2  _|-  5ab  —  b\    18a2  _  20a5  —  19^*2,    and     14a2  _  Zab 
+  20^2. 

29.  Find  the  sum  of  ^x^  —  ba^  —  bax^  +  (^a^^,  O^^  +  S^c^ 
+  4:ax^  +  2a^x,  —  llia^  +  19aa;2  —  lba%  13ax^  —  27a^x 
+  18^3,     Sa^x  —  20a^  +  12a:3^     and     SW^x  —  2x^  —  dlax- 

—  7x^  Sum,  —  W  —  a^ 

30.  Find  the  sum  of  2ab  +  12  —  x%  x^y  +  xy  +  10, 
dxy^  +  2x^y  —  xy,  bxy  +  11  +  xVy,  and  17  —  2x^y  —  x^y. 

Sum,  2ab  +  50  +  bxy  —  x^y  -\-  4a^\^y. 


ADDITION.  87 

31.  Add  x^  +  ax—ab,  ab—Vx-h^^y,  ax-\'Xy—^ab,  x^ 
-{-Vx—x,  and  xy  +  xy-\-ax. 

Sum,  2x^  4-  Sax — ^ab  +  ^xy — x. 

32.  Add  7a;i?/— 2a:V^-|-7,  Vxy  +  Sxy^-^2,  dyVx—Vyx 
—6,  9?/V^— 4i/*a;— 3,  and  l-\-7xy^—2yxK 

Sum,  l8xhj-{-3xy^  +  l. 

Scholium  2. — The  object  and  process  of  addition,  as  now  ex- 
plained, will  be  seen  to  be  identical  with  the  same  in  Arithmetic, 
except  what  grows  out  of  the  notation,  and  the  consideration  of 
positive  and  negative  quantities.  For  example,  in  the  decimal 
notation  let  it  be  required  to  add  248,  10506,  5003,  81,  and  106. 
The  units  in  the  several  numbers  are  similar  terms,  and  hence  are 
combined  into  one  :  bo  also  of  the  tens,  and  of  the  hundreds.  To 
make  this  still  more  evident,  let  u  stand  for  units,  t  for  tens,  h  for 
hundreds,  th  for  thousands,  and  t.th  for  ten  thousands.  248  is  then 
2h  +  4t  +  Su,  10506  is  lt.th-\-5h  +  Qu,  600S  is  5^^ +  3m,  81  is  8^  +  lt^, 
and  106  is  lh-\-Qu.  Writing  these  so  that  similar  terras  shall  fall 
in  the  same  column,  we  have  the  arrangement  in  the  margin. 
Whence,  adding,  we  get  the  sum.  The  process  of  carrying  has  no 
analogy  in  the  literal  notation,  since  the  relative  values  of  the  terms 

are  not  supposed  to  be  known. 
2h  -{-  4:t  -\-  8u      Again,  there  is  nothing  usually 
ILt/i  -j-  5h  -\-  6u      found  in  the  decimal  addition 

5^^  +3?^      like  positive  and  negative  quan- 

^^  i~  i^      titles.     With  these  two  excep- 
tions the  processes  are  csseu- 


U  -f  6w 


It.ih  -\-  5th  -f  9/i  -f-  4^  +  4w      tially  the  same.    The  same  may 
25944  be  said  of  addition  of  compound 


numbers. 


73.  Prop.  3. — Literal  terms,  which  are  similar  only 
with  respect  to  part  of  their  factors,  may  be  united 
into  one  term  with  a  polynomial  coefficient. 

Demonstration. — Let  it  be  required  to  add  5ax,  —2cx,  and  2mx. 
These  terms  arc  similar,  only  with  respect  to  x,  and  we  may  say  5a 


38  FUNDAMENTAL    RULES. 

times  X  and  —2c  times  x  make  (5a— 2c)  times  x,  or  (5a—2c)x.  And 
then,  5a— 2c  times  x  and  2m  times  x  make  (5a— 2c  +  2m)  times  x,  or 
(5a— 2c  +  2m)a?.     Q.  E.  D. 

EXAMPLES. 

Ex.  1.  Add  ax^,   —hx^,   —2cx^,  and  4:mx^,  with  respect 
to  x^.  Sum,  (a—h—2c-\-4:m)x^. 

2.  Add  4:xy,  Saxy,  —lOmxy,  and  cxy,  with  respect  toa;^. 

>S'wm,  (44-3a— 10m-f-c)a;?/. 

3.  Add  amx-\-2di/,  2cx~'ddy,  and  3dx-\-5y,  with  respect 
to  a;  and  ^.  /S'?<?72,  (amH-2c  +  3fi?)a;4-(5— <:?)  ?/. 

4.  Find  the  sum  of  ax^-\-iy^-\-cxy,  and  mx^~-ny^—pxy. 

Sum,  (a-\-m)  x^ -\-  (b—n)  y^ -\-  (c  —p)  xy. 

6.  Find  the  sum  of  ao(^-\-l)x^-\-cx,  and  a^of^—h^j/^—d^x. 
Sum,  {a  +  a^)7^  +  {'b  —  h^)x^'-^{c-(^)x. 

6.  Find  the  sum  of  (a  —  'b-\-c)Vx,  {a-\-'b—c)\/x,  and 
(h  +  c—a)\^x.  Sum,  {a  +  h  +  c)  Vx. 

7.  Find    the    sum     of    2ax-^y^  +  3bc  —  ^a'^x'^  -4-  3^, 
3/.a;-V^+25.  +  -f__2J,  and  ^_J  +  ^.-I-. 

Sum,  3  (a  +  A)  ^  +  52>c  +  |«-2a;-t. 

8.  Add    2x^y^  +  2x-^y^—2x^,    —  5sx^y^  +  2r^x-^y^  —  a 
+  6a;8,  3a;y— 2^»2a;-»»yt4.3«,  and  — 2a^  +  c«/. 


74.  Prop.  4. — Terms  which  have  a  common  com- 
pound, or  polynoinial  factor,  may  he  regarded  as 
similar  and  added  with  respect  to  that  factor. 

Demonstration.  5{x^—y^),  2{x^—y^)  and  — 8(2;'-^'^)  make,  when 
added  with  respect  to  {x^—y"^),  4:{x^—y'^),  for  they  are  5  +  2—3,  or  4 
times  the  same  quantity  (x'^—y^).  In  a  similar  manner  we  may 
reason  on  other  cases,     q.  e.  j>. 


ADDITION.  39 


EXAMPLES. 


Ex.  1.  Find   the   sum  of   Vx-^n—dx^,  5a^—2  Vx-i-n, 
SVx-i-n  —  Hx^,  and  10x^+8  Vx^i. 

Sum,  10  Vx  -h  w  +  6x^. 

2.  Find  the  sum  of  6a— 6(a— J)  +  7,  3a  +  12(a—b)—8, 
and  2(a—b)— 3a— 20.  Sum,  Ga— 21-^8  [a— i). 

3.  Find    the    sum    of    7(7n  +  3)— 16  (m— 3),    8(w  +  3) 
+  7(m— 3),  and  3  (m— 3)— 4  (m  +  3). 

Sum,  ll(m  +  3)  — 6(m— 3). 

4.  Find   the   sum   of  i\/a^Sh—i\/a—3h  +  i\/a^^^ 
-^\^a-3b.  Sum,  ^^a-U. 

5.  Add    3(a-c)ix-^f)-i,    — ^-~      — L==,     and 

(x-{-y^P      Vx-\-y^ 

6(a  +  c)  (.T+^2)-i.  g^^^  3(3a  +  g) 


6.  Find  the  sum  of  a(a-\-h)-\-3  va—x,  — 4a(a-}-5) 
+  1a{a~x)^,  —GaVa  —  x-\-  lla {a  +  h),  —2a{a  +  h) 
—2{a—x)^,  and  5a  (a  +  J) +14  a/^— a;. 


7.  Add  a  v^— y  +  ^a:^  +  c(a  +  a;)2,  _ Ja;i/  +  (a  +  c)(a  +  ic)2 
+  (a:— y)i,  2ia;y  +  (a— 1)  Vx—y—a  {a+xf. 

Sum,  2a  (x—y)^  +  2bxy  +  2c  (a  +  a;)^. 

8.  Show  that   Vx  +  hy  -\-  ax  —  z-\-amy-j-c  Vx  -^dz-\-y 

=  (am-^b-\-l)y-\-(c  +  l)x^-\-{d—l)z-{-ax. 

Test  Questions. — Does  addition  always  imply  an  increase? 
When  does  it  not?  When  does  it?  What  is  addition?  How  are 
similar  terms  added  ?  How  are  dissimilar  added  ?  Give  the  Rule 
for  adding  polynomials  and  demonstrate  it  ? 


40 


FUNDAMENTAL    RULEU. 


O 
Q 

a 

< 


DEFS. 


SYNOPSIS 
r  Addition. 

[Sum,  or  Amount. 

r  I.  Similar  Terms,    dem. 


PROPS. 


Cor.  l.—Siffn  of  sum. 
Cor.  2. —Sum  of  Positive 

and  Negative. 
Cor.  3  —Addition  not  al 

ways  incj'ease. 

{  Sch.— Sign  of  term. 
I  Cot.— Addition  of  Nega- 
]         tive  -  Subtraction  of 
i         Podtive. 

,  Sch.  1.— Practical  method. 
PROB.     To  add  Polynomials,  rule.  Dem.  -!  Sch.  2.— same  as  in  Led- 


^  2.  Dissimilar  Terms,    dem. 


PROPS 


\  3.  Terms  Partially  Similar. 
i  4.  Compound  Similar  Terms. 


I         mal  Notation. 


75.  Subtraction  is,  primarily,  the  process  of  taking  a 
less  quantity  from  a  greater. 

In  an  enlarged  sense,  Subtraction  comes  to  mean  taking 
one  quantity  from  another  irrespective  of  their  magnitudes. 

Subtraction  also  comprehends  all  processes  of  finding  the 
difference  between  quantities. 

The  terms  Minuend,  Subtrahend,  and  Remainder,  are  used  as  in 
Arithmetic. 

40 


BtJBTRACTION.  41 

76.  The  Difiference  between  two  quantities  is,  in  its 
primary  signification,  the  number  of  units  which  lie  between 
them  ;  or,  ii  is  what  must  he  added  to  one  in  order  to  produce 
the  other. 

When  it  is  required  to  take  one  quantity  from  another, 
t!ie  difference  is  what  must  be  added  to  the  Subtrahend  in 
order  to  produce  the  Minuend. 

Scholium. — The  most  compreliensive  and  fundamental  notion  of 
difference  is  this :  Having  reached  any  Bpecified  point  in  a  scale  of 
numbers,  or  in  estimating  magnitude,  in  what  direction,  and  how 
far  must  we  pass  to  reach  another  specified  point  in  the  scale  of 
numbers,  or  in  the  value  of  the  magnitude. 

Illustration. — When  we  ask,  "What  is  the  difference  between  3 
and  8  ? "  we  ordinarily  mean,  "  Over  how  many  units  must  we  pass 
in  reci<oning  from  3  to  8  ? "  That  is,  "  How  many  units  added  to 
3  will  make  8?" 

Let  us  use  the  following  device  to  illustrate  the  whole  subject. 
Consider  A  the  zero  point  on  the  line  BC.     Call  distances  to  the 


- 

+ 
c 

B 

A 

! 

— r 

-m, 

-U 

n 

j,j,-i, 

-1,-1. 

o.i 

+  i,  +  3,  +  4,+5, 

1      1      1 
+  6,  +  7,+8 

,+9,  +  m 

right  of  A  positive  (+ ),  and  distances  to  the  left  negative  (-).  Also 
call  reckoning  towards  the  right  positive  and  towards  the  left 
negative,  from  any  point  on  the  scale. 

1st.  The  difference  between  3  and  8,  means  either,  how  far,  and 
In  what  direction  must  we  go  to  pass  from  3  to  8  or  to  pass  from  8 
to  3  ?  In  the  first  case  we  pass  5  to  the  right,  and  say  3  from  8  is 
+  5,  understanding  that  3  and  8  are  both  positive.  But  to  pass 
from  8  to  3,  we  pass  5  towards  the  left,  and  hence  say  8  from  3 
gives  -6.  These  two  operations  are  represented  thus:  8—3  =  5, 
and  3—8  =  -5. 

2nd.  The  difference  between  -3  and  -  S  meauB  either,  how  far, 
and  in  ichat  direction  must  we  pass  to  go  from  -3  to  -8,  or  from  -8 
to  -3?     To  pass  from  -3  to  -8,  we  pass  5  towards  the  left,  and 


42  FUNDAMENTAL    RULES. 

hence  say  -3  from  -  8  gives  -5.  That  is  -8— (-3)  =  -5.  But  to 
pass  from  -8  to  -3,  we  pass  5  to  the  right,  hence  -3— (-8)  =  +5. 

3d.  The  difference  between  -3  and  +8  means  either,  how  far  and 
in  what  direction  must  we  pass  to  go  from  -3  to  +8,  or  from  +8  to 
-3?  In  the  first  case  we  get  +8— (-3)  =  +11,  since  we  pass  11 
to  the  right.  In  the  second  case  we  get  -3— (+8)  =  -11,  since  to 
go  from  +8  to  -3  we  pass  11  to  the  left. 

In  each  and  all  of  these  cases,  the  question  is,  "  What  must  be 
added  to  tlie  quantity  conceived  as  the  Subtrahend,  in  order  to 
produce  the  Minuend?"  considering,  in  each  instance,  the  number 
of  units  between  the  two  given  terms  as  the  numerical  value  of  the 
difference,  and  its  sign  as  determined  by  the  fact  as  to  whether  we 
reckon  to  the  right  (up  the  scale),  or  to  the  left  (down  the  scale), 
in  passing  from  the  Subtrahend  to  the  Minuend. 


77.  Prob. — To  perfopm  Subtraction. 

Rule. — Change  the  signs  of  each  term  in  the  subtra- 
hend frojn  +  to  — ,  or  from  —  to  -\-,  or  conceive  them 
t^  he  changed,  and  add  the  result  to  the  minuend. 

Demonstration. — Since  the  diflference  sought  is  what  must  be 
added  to  the  subtrahend  to  produce  the  minuend,  we  may  consider 
this  difference  as  made  up  of  two  parts,  one  the  subtrahend  with  its 
signs  changed,  and  the  other  the  minuend.  When  the  sum  of  these 
two  parts  is  added  to  the  subtrahend,  it  is  evident  that  the  first 
part  will  destroy  the  subtrahend,  and  the  other  part,  or  minuend, 
will  be  left. 

Thus,  to  perform  the  example : 

From  5ax—Ql)—dd—  4m 

Take  2ax  +  2'b-5d+  8m  ) 

Subtrahend  with  signs  changed,  —2ax—2b-\-5d—  8m  >• 
Minuend,  5ax—Qh-Sd—  ^m  ) 

Difference,  3aa;—8& -1-2(^—1 3m 

If  the  three  quantities  included  in  the  brace  are  added  together, 
the  sum  will  evidently  be  the  minuend.  If,  therefore,  we  add  the 
second  and  third  of  them  (that  is  the  subtrahend,  with  its  signs 
changed,  and  the  minuend)  together,  the  sum  will  be  what  is 
necessary  to  be  added  to  the  subtrahend  to  produce  the  minuend, 
and  hence  is  the  difference  sought,     q.  e.  d. 


SUBTRACTION.  43 

EXAMPLES. 

Ex.  1.  From  4:ab-\-3c^—xy  subtract  3xy—2z-\-2ab—c^. 

Model  Solution.— Writing  the  subtrahend  under  the  minuend  so 
at  similar  terms  shall 
of  combination,  I  have 


that  similar  terms  shall  fall  under  each  other,  for  the  convenience 


4a5  +  3c'  —    xi/ 
2ah—    c"  +  3a^  —  2e 
2ab  +  4c'  —  ixy  +  2z 

The  difference  sought  may  be  considered  as  consisting  of  two 
parts:  1st,  the  subtrahcml  with  its  signs  changed,  and  2nd,  the 
minuend  itself.  The  sum  of  these  two  parts  is  the  difference,  since 
it  is  what  is  necessary  to  be  added  to  the  subtrahend  to  produce 
the  minuend.  Therefore  conceiving  the  signs  of  the  subtrahend  to 
be  changed,  and  adding,  I  have  2aJ>+4:c^—4:xy-\-2z,  as  the  difference 
sought. 

A  more  detailed  Explanation  is  as  follows:  The  first  question  is, 
What  must  be  added  to  —2z  to  produce  the  corresponding  term  in 
the  minuend?  This  is  evidently  +2z,  as  there  is  no  corresponding 
term  in  the  minuend,  and  I  have  only  to  write  a  term  in  the  differ- 
ence which  will  destroy  —2z  when  added  to  it.  Passing  to  the 
next  term,  I  inquire.  What  must  be  added  to  +  3a^  to  produce 
—xyl  First,  I  must  add  —Sxi/  (the  term  with  its  sign  changed)  in 
order  to  destroy  the  +  Sxy,  and  then  I  must  add  —xy  in  order  to 
make  the  —xy  of  the  minuend.  So  in  all  I  must  add  —Sxy  and 
—xy,  or  —ixy.  —^xy  is,  therefore,  the  difference  sought.  Passing 
to  the  next  term.  What  must  be  added  to  — c'  to  make  +3c'?  I 
must  add  +c'  (to  destroy  — c')  and  +3c'  (to  make  up  the  required 
term  in  the  minuend),  or  in  all  +c'  and  +3c',  or  +4c' ;  which  is  the 
difference.  Finally,  the  term  to  be  added  to  2oib  in  order  to  make 
kcib  is  composed  of  the  two  parts  —2ab  and  +4«5,  which  make  2a5. 

It  thus  appears  that  2a5  +  4c'-4icy  +  23  is  the  difference,  since  it 
is  what  m  ust  he  added  to  tfie  subtrahend  to  produce  the  minuend. 

2.  From  ^— 311^-^  %xy -^11  take  2x-\-l^^-'2xy—b. 

Rem.,  22'-4Z.2_G. 


44  FUNDAMENTAL    RULES. 

3.  From  15fla;-f2%—6a^a;2  take  — 5«a;— 4%— 3aW 

Rem.,  20«^-f  G%— 3a2a;a. 

4.  From  8«^J-2_3^^_^15_2v'^  take  8  +  10a:z/— 3tt^J-2 

i?em.,  llflV^— 13a;?/4-7— 4Vic^. 

5.  From  ia^x^—'dc^l'^hhj^  take  8%i4-4a2.Ti_2d 

6.  From  «2—2aZ>  +  ^>2  take  ^2_j_2aJ  +  ^>2. 

^em.,   —  4ai. 

7.  From  a2_j2  take  a^—'lab—W.  Rem.y  2ab. 

8.  From    24a:/— 14m?/  +  18a;y— 14  +  27a;;22   take    17a;/ 
—  10my—4.xy-{-20xz^—S. 

9.  From   17pmx^  —  ISn^  +  lOm*  —  24  take   7pmx^—An^ 
+  10m4— 17. 

10.  Yrom  a^-^2abi-i^  take  a^—2ab-^  hi 

11.  Froma3+3a2^,4-3«JH^nakefl3_3^2j_^3«j2_53. 

1 2.  From  a:^  +  2a:%^  -\-  yi  take  a:t — 2x^y^  +  ^i 

13.  From  eVxTy  +  ^a^^  take  3  (.t  +  «/)^— 46?2a;2. 

i>?J.,  SVx  +  y  +  SaHl 

Suggestion. — Regard  \/x  +  y  as  one  quantity,  and  observe  that 
'J  (x  +  y)^  is  the  same  as  3  '\/x  +  y. 

14.  From    6a;— 3  Vxy  4-  17  (a  +  b)    take     2x  +  1l  Vxy 
+  4  («+*).  />^y.^  4a:-10A/.^  +  13(«  +  ^). 

15.  From  10— a-\-b—c-]-x  take  ft  +  i— c— a;— 90. 

16.  From  6m  +  i^n— frz  — 1  take  2?7i— 4w  +  -i«  — 2. 

Diff;  "km-^n — a  +  1. 


SUBTRACTION.  45 

Scholium  I. — When  several  polynoniials  are  to  be  combined, 
some  by  addition  and  some  by  subtraction,  it  will  be  found  expe- 
dient to  write  them  so  that  similar  terms  will  fall. under  each  other, 
writing  the  several  subtrahends  with  their  signs  changed,  and  then 
add  the  quantities  as  they  stand. 

17.  From  the  sum  of  2ay'^—2b-\'3ax,  2ay^  +  b—aXy  and 
'Sb—ay^+d—y,  subtract  3ay--{-ib--ax-\'i/. 

Resu  It,  —  2i  +  'dax  -{-'6—'Zy. 

18.  From  the  sum  of  'da^-\-xy^—2by,  6a^-\-3xy^^—3byy 
and  3xy^-\-4:a^-{-by,  subtract  2a^^xy^—5by  +  5. 

Result,  l0a^-{-8xf-{-by—b. 

19.  From  the  sum  of  a^b^—3y'^  +  6xy,  3a^l^-\-3y^—2xy, 
and  5^'  +  3aW—xy  -f-  6,  subtract  a^b'^  -f  xy—y^. 

Result,  QaW  +  ^y^+xy^^. 

20.  To  5aa;2_7c^»-f  8mi— 2c-^  add  3cb—4:C-^  +  %ax'^, 
then  subtract  IQm^ —bxy ■\-3cb—l)lc-^,  add  Qm^—l\cb  +  3n 
—iax^,  subtract  —16ax^—2m^  +  4iCb,  add  5m^-|-6c-^  and 
subtract  4:Xy—2n. 

Result,  ldax^—22cb  +  llmi  +  12c-"' ^-xy-\- 5n. 

21.  To  lcy-^-\-Sax—bb,  add  ^b—2cy"^+m,  then  sub- 
tract 5a2;— 4m  +  3  and  —3ax-\-bcy~^—Q,  add  10ax—2h 
4- 8m — 3,  and  subtract  3m—10cy~^—2m. 

Result,  li)cy~^^-^\Qax—3b  +  l2m. 

Scholium  2. — The  proficient  solves  such  examples  as  the  above, 
and,  in  fact,  all  kindred  ones,  without  re-writing  the  quantities. 
Thus,  to  obtain  the  result  in  Ex.  20,  he  looks  at  Saa:",  casts  his  eye 
along  till  he  sees  2aar'',  says  lax^ ;  then  looks  forward  to  — 4aa;',  and 
says  Zax^ ;  then  notices  —IQax^  in  a  subtrahend,  and  hence  tliinks  of 
it  as  -f  IGtfa?^,  and  says  l^ax^.  Again  taking  up  — 7e5,  he  looks 
along  noticing  the  similar  terms  and  says,  mentally,  — 4cJ,  — 7c&, 
— 18c&,  — 22c&.  Again,  he  takes  up  Ba/^I,  and  running  through 
with  the  similar  terms,  says,  —  2/ni,  4;»i,  6mi,  Umi.  And  so  for 
all  the  other  terms. 


46  FUNDAMENTAL    RULES. 

This  is  the  prcbctical  way  of  solving  such  examples  ;  and  the  pupil 
should  exercise  himself  till  he  can  write  out  the  result  ai  once.  He  can 
go  over  the  preceding  examples  thus,  and  should  also  solve,  with- 
out writing,  those  which  follow. 

22.  From  ldx^—2ax—W  take  hx^—'lax—}^, 

23.  From  20ax —^x^M  take  ^ax  +  5a;^ — d. 

24.  From  hab^W—c-\-hc-h  take  I^—^ab  +  bc. 

25.  From  ax-^—ax-'^-\-cx—d  take  a^—ax-^^—ex—'M. 

Diff.,  a  (x-^-  x^)  J^{c  +  e)x  +  d. 

Suggestion. — This  would  at  first  become  ax-^—ax^-\-cx-\-ex-\-d 
But  this  may  evidently  be  written  as  above. 

26.  From  x'^y—^  ^/xy—^ay  take  3:^:2^  +  3  (xy^—^ay. 

27.  From  the  sum  of  4o5a;— 150-f-4a;^,  hx^-\-Zax^V)^x, 
and  90— 2«a;— 12\/ic;  take  the  sum  of  %ax—'^^-\-'^x^, 
7a;i_8«a;— 70,  and  30— 4a:^,  — 2a:2  +  4ttV. 

Result,  llfl!2;  +  60— rri— 4a2ic8. 

78.  Cor.  1. —  When  a  parenthesis,  or  any  symbol  of  like 
signification  (50),  occurs  in  a  polynomial,  jrreceded  by  a  — 
sign,  and  the  parenthesis  or  equivalent  symbol  is  removed,  the 
signs  of  all  the  terms  which  were  within  must  be  changed,  since 
the  sign  —  indicates  that  the  quantity  within  the  parenthesis  is 
a  subtrahend. 

28.  Eemove  the  parenthesis  from  the  polynomial  3a^x 

-\-2i^y^ — {oa^x — 2mh-\-8b'^y^)  and  represent  the  result  in  its 
simplest  form.  Do  the  work  mentally,  writing  only  the 
result. 

EesuU,  2m'^z—2ah-eby. 

29.  Remove  the  parenthesis  from  ba—^b-\-oc—{—3a 
'\-2b—c).  EesuU,  8a— 6^4- 4c. 


SUBTRACTION.  47 

30.  Remove  the  parenthesis  from  4:a  —  5x  —  {a  —  4:x) 
-i-(x—Sa).  Result^   —5a, 

Queries. — In  Ex.  28  is  the  sign  of  5a^x  changed  ?  What  is  its 
sign  as  it  stands  in  the  parenthesis  ?    Is  the  —  sign  which  appears 

in  the  result  before  5a^x,  the  same  as  the  one  before  the  parenthesis 
in  the  example  ?  No,  What  became  of  that  before  the  parenthe- 
sis? Ans.  The  operation  which  it  indicated  having  been  per- 
formed, it  was  dropped.  In  Ex.  30,  why  are  not  the  signs  of  the 
terms  in  the  last  parenthesis  changed  ? 

79.  Cor.  2. — A  parenthesis  preceded  hy  the  —  sign  can 
be  placed  around  any  number  of  terms  by  changing  the  signs 
of  all  the  terms.  The  reaso7i  of  this  is  evident,  since  by 
removing  the  parenthesis  according  to  the  preceding  corollary, 
the  expression  would  return  to  its  original  form. 

31.  Introduce  within  a  parenthesis  the  3d,  4th  and  5th 
terms  of  the  following  expression:  6ax—2cd—8m-{-ox—2y 
-fa;— 4«. 

Result,  (jax—2cd—(8m—5x-^2y)-\-x—4:a. 

32.  Introduce  within  a  parenthesis  the  last  three  terms 
of  ixy-{-2cb—8x—5-^2b. 

Result,  4:xy-{-2cb—{Sx-j-5—2b). 

33.  Include  in  brackets  the  3d,  4th,  and  5th  terms  of 
5ax—2a^-{-3x—12ay-{-16.  Also  the  4th  and  5th.  Also 
the  2d  and  3d. 

Form  of  the  first  and  last,  bax—23^-\-{dx—12ay-\-\b), 
bax—{2x'^—3x)  —  12ay-\-\6. 

Query.— In  the  last  is  the  sign  of  2a;'  changed  ? 

34.  Prove  that  (3x—Qy)  +  (4y— 4a:)  -^  2  {x  +  2y)  =  x 
+  2y. 

35.  Prove  that  i(«  +  5—c)4-i(^  +  c—«)  =  b. 

36.  Prove  that  Qa—ib—2  (a  +  b)  =  2  (2a— 3b). 


4»  FUNDAMENTAL    RULES. 

37.  Prove  that  x^+ex^y-\-d  —  {x^-i-4:X^t/-{.l)z=2(x^y-{-l). 
*38.  Prove   that    i{a—5b-{-ic)  i- i  {5b  —  2a -]- ^c)  =  ^c 

39.  Prove  that  ^  (a^—a-\- 1 )  +  i  (2a2_«  +  2)  =  ^  (10«2 
-7«  +  10). 

40.  Prove  that  ^a  +  ib—{ia—}b)  =  b. 

41.  Prove  that    i{9—15x)—i(l2-20x)—i{4:bx-20) 

=  1— 4.T. 

42.  Prove  that  i{a-ib  +  ic)-hi  {a-ib+}c)  =  ^(90a 
—365  + 25c). 

80,  Cor.  3. —  When  sevei-al  parentheses  occur,  included  the 
one  within  the  other,  begin  the  removal  with  the  inside  one. 

43.  Remove  the  parentlieses  and  other  marks  of  aggre- 
gation from  ^a—\  —  Yc—d-\-{4:X^—\)—xy']  —  '6y\. 

Result,  4^  +  <? — c?  +  4.^2—  l—xy-\- Sy. 

44.  Show  that  «_ [J_  |c-(^-^7)j] 

=  a-[b-\c-{d-e+f)\] 
=  a-[b-\c-d+e-f\] 
=  a  —  [b—c-\-d—e-\-f] 
=  a—b  +  c — d  +  e—f. 
46.  Remove  the  marks  of  aggregation  from  the  expres- 
sion  1a—\Za—[^ia—{ba—2a)']\,    and   afteriuards  reduce 
the  result  to  its  simplest  form.     Also  combine  and  remove 
at  the  same  time.  Result,  6a. 

46.  Remove  the  marks  of  aggregation  from  a  +  2b —  \6a 
—  [3b+{8x~2-}-by-x)  +  4.a]-3b\. 

Result,  8b—a-\-'l!x—by—2. 

Scholium  3. — Terms  having  common  literal  factors  may  be 
regarded  as  similar  with  respect  to  these,  and  treated  accordingly, 
tlie  other  factors  in  each  term  being  regarded  as  coefficients. 


*  "niis,  and  the  examples  which  follow  in  this  section,  may  be  omitted  and  taken 
In  the  review  after  Fractions,  if  thought  best  by  the  teacher. 


.SUBTRACTION. 


4d 


47.  From  ax-\-hy—cz  take  my—nx-\-2z. 

Result,  {a -\-n\x-\-  (h — m)  y  —  (c-\- 2)  z. 

48.  From  ax? -\-hxy -^ cy^  take  dx^—hxy-\-Tcy^. 

Result,  (a —d)  3^  ^  (b-{-h)  xy  -\-  {c— k)  f. 

49.  From  Vx+^+Sax—12  take  4:(xi-y)^-j-b—2ax. 

Result,  5ax—3(x-^yy^—12—b. 

60.  From  a^x^—^axy-\-^a^xY  take  c^a^—Scxy  +  Gx^. 

Remit,  (a^—ci)xi—{^a—Sc)  xy-j-(4d^—Q)x^yl 

Show  that  the  second  tenii  in  the  last  result  may  be  4-  (8c— 4a). 

61.  From  Vx^  —  y^  —  2  (a  -\-  x)^  -{- 3   take  —3  Va  +  x 
4.4(avj_y2)i_i.  Result,  {a-{-x)^—3Vx^—y^-^4:. 


</> 


SYNOPSIS    FOR    REVIEW. 

Subtraction. 

PEnN.T.ONS.J,"-*' 

_.«.  j  SCHOLIUM. 

I.  uinerence.  \  illustration,   diagram. 

O  GENERAL       1  j  Sch.  1. -Both  add.  and  Bub. 

<^       PROBLEM.      |R"'e.  »EM.         1  sch.  g.-Practical  method. 

F  f  Cor.  I. — To  remove. 

S      BRACKETS.     ]  Cor.  2.— To  introduce. 
3  [  Cor.  3.— Several. 

TERMS  PARTIALLY  SIMILAR.     Sch.  3. 


Test  Questions.— What  two  answers  can  you  give  to  the  ques- 
tion, "  What  is  the  difference  between  10  and  6  ?  "  Why  do  you 
change  the  signs  of  the  subtrahend  in  subtracting?  Why  do  you 
add  the  subtrahend,  with  signs  changed,  to  the  minuend  ?  When 
do  you  change  the  signs  in  removing  a  parenthesis  ?  Wliy  ?  What 
becomes  of  the  sign  before  the  brackets  ?  In  removing  a  parenthe- 
sis preceded  by  a  —  sign,  is  the  sign  of  the  first  term  changed  as 
well  as  the  others  ?  State  in  the  briefest  manner  the  tlieory  of  sub- 
traction ?  Rejjly.  Subtraction  is  finding  the  diflFerence  between 
quantities,  that  is,  finding  what  must  be  added  to  one  quantity  to 
produce  the  other.  This  diflference  may  always  be  considered  as 
3 


60 


FUNDAMENTAL    RULES. 


consisting  of  two  parts,  one  of  which  destroys  the  subtrahend,  anc 
the  other  part  is  the  minuend  itself.  Hence,  to  perform  subtrac- 
tion, we  change  the  signs  of  the  subtrahend  to  get  that  part  of  the 
difference  which  destroys  the  subtrahend,  and  add  this  result  to 
*he  minuend,  which  is  the  other  part  of  the  difference.' 


JSL 


icatinq^ 


81,  Multiplication  is  the  process  of  finding  the  sim- 
plest expression  for  a  quantity  which  shall  be  as  many 
times  a  given  quantity,  or  such  a  part  of  that  quantity,  as 
is  represented  by  a  specified  number. 

The  quantity  to  be  multiplied  is  called  the  Multiplicand. 
The  number  by  which  we  multiply  is  called  the  Multiplier. 
Taken  together,  the  multiplier  and  multiplicand  are  called 
Factors.    The  result  is  the  Product, 

82,  Cor.  1. — The  multiplier  must  ahvays  he  conceived  as 
an  abstract  yiumber,  since  it  shows  how  many  times  the  mul- 
tiplicand is  to  be  taken. 

Thus,  to  propose  to  multiply  $13  by  $5  is  absurd.  We  can  under- 
stand that  5  times  $12  is  $60 ;  but  what  is  meant  by  5  dollars 
times? 

83,  Cor.  2. —  TJie  product  is  always  of  the  same  kind  as 
the  multiplicand. 

84,  Scholium. — It  is  frequently  convenient  in  practice  to  speak 
of  the  multiplier  as  positive  or  negative,  although,  literally  under- 
stood, this  is  a  contradiction  of  Cor.  1,  which  requires  the  multi- 
plier to  be  conceived  as  mere  number.     In  a  strict  analysis,  the 


MULTIPLICATIOlf.  51 

multiplier  in  such  cases  is  to  be  cousidered,  first,  without  reference 
to  its  sign,  i.  «.,  as  abstract,  and  then  the  sign  is  to  be  interpreted 
as  indicating  what  is  to  be  done  with  the  product,  when  it  is  taken 
in  connection  with  other  quantities. 


So.  Prop.  1. — The  product  of  several  factors  is  the 
same  in  whatever  order  they  are  taJcen. 

Demonstration. — 1st.     ax6,  is  a  taken  &  times,  or  a-\-a  +  a-^a 

+  a to  b  terms.     Now,  if  we  take  1  unit  from  each  term 

(each  a),  we  shall  get  h  units ;  and  this  process  can  be  repeated  a 
times,  giving  a  times  6,  or  6 x a.  .*.  axb  =  bx a. 

2nd.  When  there  are  more  than  two  factors,  as  abc.  We  have 
shown  that  ab  =  ba.  Now  call  this  product  /»,  whence  abc  =  mc. 
But  by  part  1st,  mc  =  cm.  .*.  abc  =  bac  =  cab  =  cba.  In  like  man- 
ner we  may  show  that  the  product  of  any  number  of  factors  is  the 
same  in  whatever  order  they  are  taken,     q.  e.  d. 

Scholium- — If  the  multiplicand  is  concrete,  the  reasoning  is  still 

the  same.     Thus  $a  x  &  =  $a  +  $a  +  $«  +  |a, -  etc.  to  b 

terms.  Now  take  $1  from  each  of  the  terms  of  %a  each,  and  we 
have  %b  ;  and  this  process  can  be  repeated  a  times,  giving  %b  x  a.  .-. 
|a  X  ft  =  %bxa.     Notice  that  in  each  case  the  multiplier  is  abstract. 

SG,  Prop.  2. —  When  two  factors  have  the  same 
sign  their  prod^uet  is  positive ;  when  they  have  d,if- 
ferent  signs  their  product  is  negative. 

Demonstration.— 1st.  Let  the«factoi*s  be  +a  and  +&.  Consider- 
ing a  as  the  multiplier,  we  are  to  take  +&,  atimes,  which  gives  +«ft, 
a  being  considered  as  abstract  in  the  operation,  and  the  product, 
+  aft,  being  of  the  same  kind  as  the  multiplicand  ;  that  is,  positive. 
Now,  when  the  product,  +  a&,  is  taken  in  connection  with  other 
quantities,  the  sign  +  of  the  multiplier,  a,  shows  that  it  is  to  be 
added;  that  is,  written  with  its  sign  unchanged.  .".  (  +  Z»)x(  +  a) 
=  -\-ab. 

2nd.  Let  the  factors  be  —a  and  —b.  Considering  a  as  the  mul- 
tiplier, we  are  to  take  —5,  a  times,  which  gives  —ab,  a  being  con- 
sidered as  abstract  in  the  operation,  and  the  product,  —ab,  being  of 
the  same  kind  as  the  multiplicand  ;  that  is,  negative.  Now,  when 
this  product,  —aft,  is  taken  in  connection  with  other  quantities,  the 


52  FUNDAMENTAL    RULES. 

sign  —  of  the  multiplier  shows  that  it  is  to  be  subtracted;  that  is, 
written  with  its  sign  changed.  .-.  (—5)  x  (—a)  =  +a6. 

3d.  Let  the  factors  he  —a  and  +5.  Considering  a  as  the  multi- 
plier, we  are  to  take  +5,  a  times,  which  gives  -\-ab,  a  being  consid- 
ered as  abstract  in  the  operation,  and  the  product,  +ab,  being  of 
the  same  kind  as  the  multiplicand;  that  is,  positive.  Now,  when 
this  product,  -\-ab,  is  taken  in  connection  with  other  quantities,  the 
sign  —  of  the  multiplier  shows  that  it  is  to  be  subtracted;  that  is, 
written  with  its  sign  changed.  .-.  (-I-&)  x  {—a)  =  —ah. 

4th.  Let  the  factors  be  +a  and  — &.  Considering  a  as  the  mul- 
tiplier, we  are  to  take  —5,  a  times,  which  gives  —ab,  a  being 
considered  as  abstract  in  the  operation,  and  the  product,  —aft, 
being  of  the  same  kind  as  the  multiplicand ;  that  is,  negative. 
Now,  when  this  product,  —db,  is  taken  in  connection  M'ith  other 
quantities,  the  sign  +  of  the  multiplier  shows  that  it  is  to  be 
added;  that  is,  written  with  its  own  sign.   .-.  (—5)  x(  +  a)  =  —ah. 

Q.  E.  D. 

87.  Cor.  1. — The  product  of  any  number  of  positive  fac- 
tors is  positive. 

Thus  (+a)  X  {  +  h)  X  (  +  c)  x  (  +  d)  —  ahcd,  since  (  +  a)  x  (-f  &) 
=  +ah,  which,  in  turn  multiplied  by  -\-c,  gives  i-dbc,  etc. 

88,  Cor.  2. — The  product  of  an  even  number  of  negative 
factors  is  positive ;  since  we  can  multiply  them  two  and  two, 
thus  obtaining  positive  products,  which  positive  products  multi- 
plied together  make  the  complete  product  positive. 

Thus  (-a)  X  i-h)  x  {-(^  x{—d)x  (-e)  x  (— /)  =  (  +  ab)  x(  +  cd) 
X  ( -1-  ef)  =  +  ahcdef  or  ahcdef. 

89*  Cor.  3. —  The  product  of  an  odd  number  of  negative 
factors  is  negative  ;  since,  by  the  last  corollary,  the  product  of 
all  but  one  (an  even  number)  of  such  factors  is  positive,  and 
then  this  multiplied  by  the  remaining  negative  factor  gives 
( -f-)  X  ( — ),  and  hence  is  negative. 


90,  Prop.  3. — The  product  of  two  or  more  factors 
consisting  of  the  same  quantity  affected  with  expo- 
nents, is  the  com^mon  quantity  with  an  exponent 


MULTIPLICATIOK.  53 

equal  to  the  sum  of  the  exponents  of  the  factors. 
That  is,  rt*^*  X  a'*  =  «''*'■";  or  ft'^^-a'^.a*  ==  a^+"+*,  etc., 
whether  the  exponents  are  integral  or  fractional,  positive 
or  negative. 

Demonstration. — 1st.  When  the  exponents  are  positive  integers. 
Let  it  be  required  to  multiply  a^  by  a",  a*  =  aaa,  and  a^  =  aa. 
.'.  a^xa'^  =  cum  •  aa  =  a'.  That  is,  there  are  three  factors  each  a,  in 
u^,  and  tico  like  factors  in  a- ;  and,  as  the  product  consists  of  all  the 
factors  in  both  multiplier  and  multiplicand,  it  will  contain  Jke 
factors  each  a,  and  hence  is  a\ 

In  general :  To  multiply  a'"  by  a"  and  a*,    a*"  =  a/m to  m 

factors,  ft"  =  aaaaa to  n  factors,  and  a*  -—  aaaaa to  « 

factors.  Hence  the  product,  being  composed  of  all  the  factors  in  the 
quantities  to  be  multiplied  together,  contains  m  -\-n-\-s  factors  each  a, 
which  is  represented  a'"+"+».  Since  it  is  evident  that  this  reasoning 
can  be  extended  to  any  number  of  factors;  the  proposition  is 
proved  in  ^e  case  of  positive  integral  exponents. 

2nd.  When  the  exponents  are  positive  fractions.  Let  it  be  required 
to  multiply  64^  by  64^  Now  64^  =  4-4,  /.  «.,  2  of  the  3  equal 
factors  which  make  64.  In  like  manner  64^  is  2  ■  2  •  2  2  •  2.  And 
since  4-4  is  2  ^^  2- 2,  64^  x  64^  =  2-2- 2- 2  x  2- 2- 2  2- 2  =  2»,  or  9 
of  the  6  equal  factors  into  which  64  is  resolvable,  and  may  be 
represented  64«  .-.  64tx64t  =  2"  =  64^  =  64t  +  t. 

m  em 

In  general,  let  a»  be  multiplied  by  d^.  a*  means  m  of  the  n 
equal  factors  composing  a.  Now,  if  each  of  these  n  factors  be 
resolved  into  h  factors,  a  will  be  resolved  into  hn  factors,  and  to 

make  the  quantity  a»  we  shall  have  to  take  hm  instead  of  m  factors. 

>H  hm  0 

Hence  a"*  =  a^.    In     like   manner    a^   may   be   shown   equal   to 

c*  m  c  hm  en 

a^ ;  whence  a"  x  a^  =  a^  x  a^.  This  now  signifies  that  a  is  to  be 
resolved  into  hn  factors,  and  hn  +  em  of  them  taken  to  form  the 

me  bm  en  bm  +  en  me 

product.  .  •.  a*xa^  =  a'^xc^  =  a"~J»",  or  a »  ^.  Finally,  as  it  is 
evident  that  this  reasoning  can  be  extended  to  any  number  of  fac- 
tors, the  proposition  is  proved  in  the  case  of  positive  fractional 
exponents. 


54  FUKDAMENTAL    RULES. 

3d.   When  the  exponents  are  negative.     Let  it  be  required  to  tntilti- 

ply  2-  by  2-      2-  is  1,  and  2"^  is  t    .-.  2^x2-  =  \,>^\, 

1  «  . 

=  -  ,  or  2-^ 

2^' 

In   general,   «-™  x  «-"  =  a-^-^.      For  a-"*  =  — ,  and  a-"  = 

(43).      Whence  «-'"  x  a-»  =  —  x  —  = =  ^—  by  the  pra- 

ceding  parts  of  this  demonstration;  and  by  (43)  =  «-'»-». 

This  reasoning  may  also  be  extended  to  any  number  of  factors. 

Q.  E.  D. 

EXAMPLES. 
Ex.  1.  Prove  as  above  that  81^x81^  =  81^  and  that 

8lV-  =:  8ll 

2.  Prove  that  m<^  x  m*  =  m<^+*.  ^ 

3.  Prove  that  16"^  x  16"^  =  16"^. 

4.  Prove  that  25"^  x  25*  is  1. 

6.  Prove  that  a~^  x  a^  is  a. 

Scholium. — The  student  must  be  careful  to  notice  the  difference 
between  the  signification  of  a  fraction  used  as  an  exponent^  and  its 
common  signification.  Thus  f  used  as  an  exponent  signifies  that  a 
number  is  resolved  into  3  equal /actors,  and  the  product  of  2  of  them 
taken ;  whereas  f  used  as  a  common  fraction  signifies  that  a  quantity 
is  to  be  separated  into  3  equal  pa/rts^  and  the  sum  of  two  of  them 
taken. 


91.  Prob. — To  multiply  Monomials. 

Rule. — Multiply  the  numerical  coefficients,  and 
to  this  product  affix  the  letters  of  all  the  factors, 
affecting  each  with  an  exponent  equal  to  the  sum 
of  all  the  exponents  of  that  letter  in  all  the  factors. 


MULTIPLIOATIOK.  56 

The  sign  of  the  product  will  he  +  except  when  there 
is  an  odd  nwinher  of  negative  factors;  in  which 
case  it  will  he  —. 

Demonstration. — Tliis  rule  is  but  an  application  of  the  preced- 
ing principles.  Since  the  product  is  comi^osed  of  all  the  factors  oi' 
the  given  factors,  and  the  order  of  arrangement  of  the  factors  in  the? 
product  does  not  affect  its  value,  we  can  write  the  product,  putting 
the  continued  product  of  the  numerical  factors  first,  and  then 
grouping  the  literal  factors  so  that  like  letters  shall  come  together. 
Finally,  performing  the  operations  indicated,  by  multiplying  the 
numerical  factors  as  in  the  decimal  notation,  and  the  like  literal 
factors  by  adding  the  exponents,  the  product  is  completed. 


EXAMPLES. 

Ex.  1.  Multiply  together  Sa^bx,  'Zch^y,  and  5ach!, 

Model  Solution.— Since  the  product  must  contain  all  the  factors 
of  the  given  factors,  and  the  order  of  arrangement  is  immaterial 
(85),  'da^hx  X  2cb^y  x  5ac^x  =  3  •  2  •  5  x  a^a  xW  xcc^  xxxxy,  which, 
by  performing  the  operations  indicated,  becomes  30a^&Wy. 

2.  Multiply  together  Sabxy,  2a^ba:^f  lOcx^,  4?/^,  and  a. 

Prod.,  UQxi^hx^if. 

3.  Multiply  together  bax^,  —'^bij,  —3a^c,  ax^,  and  —2y\ 

Prod.,   —eOa^bcx^f. 

4.  Multiply  together  mz'*,  2rtvh?,  —3ax^,  —  5m«,  and  4ta^x^. 

Prod.y  120a%3+*:c5+r+n 

6.  What  is  the  product  of  —2d^d^  —I0ac%  —d% 
— -ia'wx",  and  —  c^.  Ans.,  —SOa^^h'>^^'^-^^d^xi^-^K 

6.  Multiply  3a;i  by  2a;i  Prod.,  6a:i 

7.  Multiply  60at  by  Sa*.  Prod,,  480«i 

8.  Multiply  3a^b^  by  —laW'  Prod.,   —2\aUh 


66  J-tJNDAMENTAL    RULES. 


9.  Multiply  6ah^  by  '^a^-^c.  Prod.,  35ah. 

10.  Multiply  10a2  by  3a~^.  Prod.,  30. 


1     .  -  _         ,  H-2" 


11.  Multiply  Sy""  by  62/2.  p^od.,  ISy 

12.  Multiply  13a;^  by  5ic^.  Prod.,  65a: ««. 

13.  Multiply  — 50a^  by  —4ta^.  Prod.,  200ak 

14.  Multiply  309a^/;^  by  9«'^^>^.  Pro^.,  2'7Sla^b^+'^. 

15.  Multiply  —6x-^y-'^  by  —  Sa:^^^.  Pro^.,  ISa:^/. 

16.  Multiply  x^  by  ic~^.  Prorf.,  a;i 

17.  Multiply  .^  by  ar-^.  Prod.,  1. 

18.  Multiply  «2j-c  by  a^bc.  Prod.,  a\ 

19.  Multiply  Va^  by  «%.  Prod,  a^x^. 

20.  Multiply  v^2  by  ^a\  Prod.,  a'^\ 


92.  Prob. — To  multiply  two  factors  together  when  one 
or  both  are  polynomials. 

Rule. — Multiply  each  term  of  the  multiplicand  hy 
each  term  of  the  multiplier,  and  add  the  products. 

Demonstration. — Thus,  if  any  quantity  is  to  be  multiplied  by  a 
+  &— c,  if  we  take  it  a  times  (/.  e.  multiply  by  a),  then  &  times,  and 
add  the  results,  we  have  taken  it  a +  5  times.  But  this  is  taking  it 
c  too  many  times,  as  the  multiplier  required  it  to  be  taken  fl^  +  J 
minus  c  times.  Hence  we  must  multiply  by  c,  and  subtract  this 
product  from  the  sum  of  the  other  two.  Now  to  subtract  this  pro- 
duct is  simply  to  add  it  with  its  signs  changed  (YV).  But,  giving 
the  —  sign  of  c  its  effect  as  we  multiply,  will  change  the  signs  of  the 
product,   and  we   can   add  the  partial  products  as  they  stand. 

Q.  E.  D. 


MULTIPLICATION.  5? 

EXAMPLES. 

Ex.  1.   Multiply  2a—Sb-\-4:C  by  'Sa-\-2b—5c. 

Model  Solution.— Writing  the  multiplier  under  the  multiplicand, 
IS  a  matter  of  convenience,  I  have 

2«   —  35    +    4c 
da   +  2b    —    5c 


6a«  —  Qab  -\-  12ac 

+  iab  _  6&'  +    8&C 

—  lOar,  +  15bc  -  ZOc" 

6a'  -5ab  +    2ac  —  65'  +  235c  —  20c' 

Now  taking  the  multiplicand  8a  times,  I  have  6a'— 9a5+12ac. 
Taking  it  26  times,  I  have  4«5— 65'  +  85c.  I  have  thus  taken  it  too 
many  times,  by  5c  times.  Hence  I  am  to  take  it  5c  times,  and  then 
subtract  this  partial  product  from  the  others.  Therefore  I  multiply 
by  5c,  and  change  the  signs  as  I  proceed,  and  finally  add  the  three 
partial  products.     I  thus  obtain  3a +  26— 5c  times  the  multiplicand. 

2.  Multiply  x-\-y  by  j'-^j/.  Prod.,  7^-\-2xy  +  i/. 

3.  Multiply  5a;-|-4y  by  Sx—2y. 

Prod.,  15a^-\-2xy—Sy\ 

4.  Multiply  x^-}-xy—y^  by  x—y. 

Prod.,  a^—2xy^+y\ 

5.  Multiply  2a^-Uy  by  2^3-3^1 

Prod.,  ^ac^ — 6bc^y — Qac'^y'^  +  %f. 

6.  Multiply  a^-i- 52^ ^2_^^_^^_^^  l^y  ^_|_^_|_^, 

7.  Multiply  «H3«2a;  +  3«:r2  4-2:3  by  a^—Za^x-Jr^a3?—7?. 

Prod.,  a«—3a*x^-\-3ah.^—xf^. 

8.  Multiply  «2_^  by  a^-\-b^. 

I  n  m 

9.  Multiply   together  (a  +  b),  (a^-\-ab-\-l^),  (a—b),  and 
(a^^ab  +  l^). 


68  FUNDAMENTAL  RULES. 

Suggestion. — Perform  this  first  by  multiplying  together  I  and  II, 
and  then  III  and  IV,  and  taking  the  product  of  these  products. 
2d,  By  multiplying  the  product  of  I  and  III  into  the  product  of  II 
and  IV.  3d,  By  multiplying  the  product  of  I  and  IV  by  the  pro 
duct  of  II  and  III.  Result  in  each  case,  a^—¥. 

10.  Multiply  a^—Q^  by  7n^—n\ 

Prod. ,  a^m^ — rn^x^ — ahi^  -\-  n '^oi?, 

11.  Multiply  2a2+ 2a +  5  by  a^—a. 

Prod.,  2«4  +  3a2— 5«. 

12.  Multiply  2s;3^4^2_^8a;  +  16  by  3a;— 6. 

Prod.,  6^^4—96 

13.  Multiply  fl^ -I- &—C  by  m—n. 

Prod.,  am — an-\-hm — h7i — cm-\-cn. 

14.  Multiply  x:^-\-x^y'^-^y^  by  x^—y^.       Prod.,  x^—y^. 

15.  Multiply  a;2— 4a;  +  16  by  x  +  b. 

Prod.,  a:3_f_a;8— 4a;H-80. 

16.  Multiply  a^—c^y-\-c^y'^—ay^-\-y^  by  a-\-y. 

Prod.,  a^-\-y\ 

17.  Multiply  a;2— 50a;— 100  by  a; +  2. 

Prod.,  a;3_48a;2_200a;— 200. 

18.  Multiply  2aj8  +  3a;— 1  by  2a;2— 3a;4-l. 

Prod.,  4a;*— 9a;2  +  6a;— 1. 

19.  Multiply  x^—h^  by  x^  +  h. 

Prod.,  x^—b^x^-^-bx^—iK 

20.  Multiply  a^—h~^  by  a^—b. 

Prod.,  d—a^-^—a^b  +  bi. 

21.  Multiply  2a-i-db^  by  2^-^  +  35^. 

Prod.,  4a-i— 9^1 


MULTIPLICATIOK.  5d 

22.  Multiply  x^  +  x^y-^-^-xiy-^  +  y""^  by  x^—y'^. 

Prod.,  x^—y~\ 

23.  Multiply  (^■\-b"'  by  a^'  +  b**. 

Prod.,  a'«+^  +  «"^>»»4-a'"Z>«4-^>'w+n. 

24.  Multiply  a^a-f-ys  by  x^—yK  Prod.,  x^—y\ 

25.  Multiply  ?/i5  +  ?/iW  +  w^  by  w^ — w^. 

Prod.,  m^?i, 

26.  Multiply  Sa'^-i— 2^>'^  by  %a—W. 

Prod.,  6a"»— 4aJ^-2— 9a'^i*2_|_6^n, 

27.  Multiply  ^aP^Plr^P  +  a-'^PbP  by  da'^l^P—a^Pb-^P. 

Prod.,  6a^-P  -f-  Sa'^^p j4p  _  2a^ + s/^^^-^Si)  _  aPb-P. 

OS,  Definition. — When  an  indicated  operation  is  performed  the 
expression  is  said  to  be  expanded. 

28.  Expand    (a  +  b){a—b);    also   (x  +  y)  i'^'^  +  y)  ;    also 
(x—a){3>^-\-ax-{-a^);  also  {?n  +  n)(7n-\-n)  —  {m  —  n)(m  —  n). 

Last  result,  4mw. 


THREE     IMPORTANT    THEOREMS. 

94,    Theorem.  —  The  square   of  the   sum  of  two 

quantities  is  equal  to  the  square  of  the  first,  plus 
twice  the  product  of  the  two,  plus  the  square  of  the 
second. 

Demonstration. — Let  x  be  any  one  quantity  and  y  any  other. 
The  sum  is  a;  +  y ;  and  the  square  is,  the  square  of  the  first,  x^,  plus 
twice  the  product  of  the  two,  2xy,  plus  the  square  of  the  second, 
y\  That  is  {x-\-yy  =z  x" ^2xy^-y\  For  {x  +  yf  ^  {x+y){x-\-y) 
which  expanded  becomes  a;'  +  ^xy  +  y"^.    Q.  e.  d. 


60  I-UNBAMEKTAL    RULES. 

EXAMPLES.' 
Ex.  1.  What  is  the  square  of  2a^i-3x^? 

Model  Solution. — This  is  the  sum  of  the  two  quantities  2a'  and 
3a;^,  and  hence  by  the  theorem  the  square  equals  the  square  of  the 
first,  4a\  plus  twice  the  product  of  the  two,  12aV,  plus  the  square 
of  the  second,  9x\    .  •.  {2a''  +  Zxy  =  4a*  +  l^a'x"  +  9^*. 

Scholium. — The  pupil  should  give  mentally  the  squares  of  the 
following  expressions : 

2.  Square  4a^  +  2a:.  Resnlt,  16a  +  16a^x-\-4:X^ 

3.  Square  x~^-^xy.  Result,  x~^-\-^x^y-[-x?y^. 

4.  Square  hrr^^-UW.       Result,  ^ha-^  +  ^OaV^-^^a^K 

5.  Square  ^h-'^  +  ^x-^b-^^ 

Result,  \a%'^-\-^ab-^x-^  +  ^x-'ib-\ 


OS,  Theorem. — The  square  of  the  difference  of  two 
quantities  is  equal  to  the  square  of  the  first,  minus 
twice  the  product  of  the  first  by  the  second,  plus  the 
square  of  the  second. 

Demonstration. — Let  x  and  y  be  any  two  quantities.  The  dif- 
ference is  x—y.  Now  {x—yy  —  (x—y)  {x—y)  which  expanded 
gives  x^  —  2xy  +  y'^.     q.  e.  d. 

'examples. 
Ex.  1.   Square  2x—Sy.  Result,  4:X^—12xy-\-9yK 

2.  Square  x~^y—2y^.        -  Result,  x'^y^—4:X~^y^-i-4:y\ 

3.  Square  2«"— 3^  "         Result,  4«"— 12«"6  "-I-9J  ". 

4.  Square  m-P—n-i.    Result,  m-^—2m-Pn-^-\-n-^. 

5.  Square  3a  "'—2b  ". 

Result,  9a  '»— 12a  '"b  »H-4Z>    «. 


MULTIPLICATIOK.  61 

90.  Theorem. — The  product  of  the  sum  and  dif- 
ference of  two  quantities  is  equal  to  the  difference 
of  their  squares. 

Demonstration. — Let  x  and  y  be  any  two  quantities.  Their  sum 
isa;+y,  and  their  difference  is  x—y.  Now  {x  +  y)  multiplied  by 
(x—y)  gives,  by  actual  mulplication,  ar^— y^  or  the  diflference  of  the 
squares  of  the  two  quantities,     q.  e.  d. 

EXAMPLES. 

Ex.  1.  Find  the  product  of  2a^-\-db  and  2a^—db. 

Model  Solution. — Here  I  have  the  sum  of  the  two  quantities  2a'' 
and  3ft,  to  be  multiplied  by  their  diflference.  Now  (2a' +  36)  x  (2«' 
—  3J)  is,  by  the  theorem,  the  square  of  2a^  or  4a*,  minus  the  square 
of  the  second,  or  96^     .  •.  (2a«  +  35)  (2a'  -  36)  =  4a*  -  96' 

2.  Find  the  product  of  «-f-26  and  a— 2b. 

Prod.,  a^—4:tfi. 

3.  Find  the  product  of  2a-\-3b  and  2a— 3b. 

Prod,,  4fl2-9*2. 

4.  Find  the  product  of  1a-\-2b  and  7a— 2b. 

Prod.,  49«2_4J2. 

6.  Find  the  product  of  6a^-{-6b^  by  5a^—6b^. 

Prod.,  25a«— 36Z»*. 

6.  Find  the  product  of  m*  +  w^  and  m^—n^. 

Prod.,  m—n. 

7.  Find  the  product  of  2^  +  3^?/^  and  2W— SV- 

Prod.,  2x—3y. 

8.  Find  the  product  of  3a^lfi+2aH^  and  3aW—2aH^. 

Prod.,  9a4J«-4ajt. 


62 


FUNDAMENTAL    RULES. 


SYNOPSIS    FOR    REVIEW. 


< 
O 


UJ  I- 

o  a. 

z  O 

=3  QC 

Li.  0. 


r  Multiplication. 
Multiplier. 
Multiplicand. 
Product. 
Factors. 


COR. 


'  I.  Order  of  Factors,     dem. 


1.  Multiplier,  abstract. 

2.  Product  like  multiplicand. 
ScH. — How  the  sign  of  a  mul- 
tiplier is  to  be  understood. 

1.  Two  factors. 

2.  More  than  two. 

ScH.— Multiplicand  concrete. 

+  ^   +iGive+. 


2.  Law  of  Signs. 


DEM. 


I  COR. 


X    

X      + 


Give 


Positive  Factors. 

Even  No.  Negative  Factors. 
Odd  No.  Negative  Factors. 


r  1.  Positive  Integer. 
3.  Laws  of  Exponents,     dem.  \  ^-  Positive  Fraction. 

j  3.  Negative  Integer. 
I       Negative  Fraction. 


TO  PERFORM 
MULTIPLICATION. 

THREE 
IMPORTANT 
THEOREMS. 


Prob.  I.     WHAT?    Rule.    Demonstration. 
Prob.  2.    WHAT?    Rule.    Demonstration. 

r  I.    Square  of  Sum.  Demonstration. 

^   2.   Square  of  Difference.  Demonstration. 

13.  Product  of  Sum  and  Difference 


Dem. 


Test  Questions. — State  and  demonstrate  the  law  of  the  signs  in 
multiplication.  How  are  expressions  consisting  of  the  same  quan- 
tity affected  by  exponents,  multiplied  ?  How  many  cases  arise  'i 
Demonstrate  each.  What  is  the  difference  between  the  square  of 
the  sum,  the  square  of  the  difference,  and  the  product  of  the  sum 
and  difference  of  two  quantities?  Demonstrate.  What  is  the 
product  of  «^  and  a*  ?     Prove  it.     What  of  a^  and  a-"^  ?     Prove  it. 

What  of  «l  and  «t  ?     Prove  it.     What  of  a"'  and  a-"''  ?    Prove  it. 


DIVISION.  63 


.97.  Division  is  the  process  of  finding  how  many  times 
one  quantity  is  contained  in  another. 

The  Dividend  is  the  quantity  to  be  divided.  The 
Divisor  is  the  quantity  by  which  we  divide.  The  Quo- 
tient is  the  result,  which  shows  how  many  times  the  divisor 
is  contained  in  the  dividend.  The  Remainder  is  what  is 
left  of  the  dividend  after  the  integral  part  of  the  quotient 
is  produced. 

08.  Scholium  I. — Division  is  the  converse  of  multiplication. 
Since  a  product  consists  of  (contains)  as  many  times  the  multipli- 
cand as  tLere  are  units  in  the  multiplier,  the  multiplier  shows  how 
many  times  the  multiplicand  is  contained  in  the  product.  The 
product,  therefore,  corresponds  to  the  dividend,  the  multiplicand 
to  the  divisor,  and  the  multiplier  to  the  quotient.  But,  as  in  mul- 
tiplication, multiplicand  and  multiplier  may  change  places  without 
affecting  the  product,  either  of  them  may  be  considered  as  divisor 
and  the  other  as  quotient,  the  product  being  the  dividend. 

00.  Scholium  2.— In  accordance  with  the  last  scholium,  the 
problem  of  division  may  be  stated:  Given  the  product  of  two  fae^ 
tors  and  one  of  the  factctrs,  to  find  the  other  ;  and  the  sufficient  reason 
for  any  quotient  is,  that  multiplied  h/  the  divisor  it  given  the  dividend. 

100.  Cor.  1. — Dividend  and  divisor  may  both  he  multi- 
plied or  both  he  divided  hy  the  sam£  numher  without  affecting 
the  quotient. 


64  FUNDAMENTAL    RULES. 

10 !•  Cor.  2. — If  the  dividend  he  multiplied  or  divided  hy 
any  number^  while  the  divisor  remains  the  same,  the  quotient  is 
multiplied  or  divided  by  the  same. 

102.  Cor.  3. — If  the  divisor  be  multiplied  by  any  number 
while  the  dividend  remains  the  same,  the  quotient  is  divided  hy 
that  number  ;  but  if  the  divisor  be  divided,  the  quotient  is  mul- 
tiplied. 

103.  Cor.  4. — The  sum  of  the  quotients  of  two  or  more 
quantities  divided  hy  a  common  divisor,  is  the  same  as  the 
quotient  of  the  sum  of  the  quantities  divided  by  the  same 
divisor. 

104.  Cor.  5.— The  difference  of  the  quotients  of  two  quan- 
tities divided  hy  a  common  divisor,  is  the  same  as  the  quotient 
of  the  difference  divided  by  the  same  divisor. 

These  corollaries  are  all  immediate  consequences  of  the  notion  of 
division.  They  need  no  demmistration^  but  it  is  well  that  they  be 
fully  illustrated.     Thus  Cor.  1,  may  be  illustrated  as  follows: 

If  a  given  number  of  apples  are  divided  among  any  number  of 
boys,  each  boy  will  receive  just  the  same  number  as  if  twice  or 
thrice  as  many  were  divided  among  twice  or  thrice  as  many  boys, 
or  as  if  1^  or  ^  as  many  were  divided  among  §  or  ^  as  many  boys. 

Cor.  4  may  be  illustrated  thus :  As  3  is  contained  in  8  four 
times  and  in  6  three  times,  it  is  contained  in  8  +  Q  four  +  three,  or  7 
times. 

105.  Cancellation  is  the  striking  out  of  a  factor  com- 
mon to  both  dividend  and  divisor,  and  does  not  affect  the 
quotient,  as  appears  from  (100). 


106,  Lemma  l.-^When  the  dividend  is  positive  the  quo- 
tient has  the  same  sign  as  the  divisor  ;  but  when  the  dividend  is 
negative,  the  quotient  has  an  opposite  sign  from  the  divisor. 

Demonstration. — This  proposition  is  a  direct  consequence  of  the 
law  of  the  signs  in  multiplication,  since  the  dividend  corresponds 
to  the  product,  and  a  positive  product  arises  from  like  signs  in  the 
factors,  and  a  negative  product,  from  unlike  signs. 


DIVISION.  65 

Scholium. — In  both  multiplication  and  division,  Like  signs  give 
4-,  and  Unlike  signs  —  resfulls. 

107.  Lemma  2. —  When  the  dividend  and  divisor  consist 
of  the  same  quantity  affected  by  exponents,  the  quotient  is  the 
common  quantity  ivith  an  exponent  equal  to  the  exponent  in  the 
dividend,  minus  that  in  the  divisor.  That  is,  a^^-r-a"' 
_  (jm-n^  whether  la  and  n  be  integi*al  or  fractional,  posi- 
tive or  negative. 

Demonstration. — This  is  an  immediate  consequence  of  the  law 
of  exponents  in  multiplication,  since,  in  the  corresponding  case  the 
exponent  of  the  product  was  found  to  be  the  sum  of  the  exponents 
of  the  factors.  Now,  as  the  dividend  is  the  product  of  the  divisor 
and  quotient,  it  follows  that  the  exponent  of  the  quotient  is  the 
exponent  of  the  dividend  minus  that  of  the  divisor,     q.  e.  d. 

EXAMPLES. 

Ex.  1.  Divide  a^  by  a\ 

Model  Solution.  «^~a^  =  a*;  since  a*  x  a' =  «^,  and  division 
is  finding  a  factor  which  multiplied  into  the  divisor  produces  the 
dividend  (99). 

2.  Divide  in^  by  m^.  Quot.,  m^. 

3.  Divide  w"  by  n~K  Quot.,  tv"    ,  or  w  *»  . 


4. 

Divide  (ad)*"*  by  (abf. 

Quot., 

(a J)'"  ",  or  (ab)  -    . 

6. 

Divide  d^  by  a*. 

Quot.,  a-\  or  ^ 

6. 

Divide  a~^  by  a*. 

Quot.,  «-«,  or  - 

7. 

Divide  x~^  by  ar^. 

Quot.,  x^. 

8. 

Divide  ar-a  by  x"^. 

Quot.,  x"^,  or  — j 
x^ 

108*  Cor.  1.  —  Any  quantity  with  an  exponent  0  is  1, 
smce  it  may  he  considered  as  arising  from  dividing  a  quantity 
by  itself 

Thus  x°  may  be  considered  as  if'-^oT  which  is  1,  because  divi- 
dend and  divisor  are  equal.     But  by  the  law  of  exponents  af"H-af" 


66  FUNDAMEKTAL    RULES. 

=  x'^. .-.  x'^  =  1.    In  like  manner  the  significance  of  any  quantity 

with  an  exponent  0  may  be  explained.     6°  =  1,  for  ^2  =  1,  and  also 

5 

~  =  5\     .'.  5°  =  1. 

109,  Cor.  2. — Negative  exponents  arise  from  division 
vjhen  there  are  more  factors  of  any  number  in  the  divisor  than 
in  the  dividend. 

Thus    a«-^a^  =  a'-^   (lOV)  =  a-^       Again,   ««-=-a^  = -' =  1 . 

or      a^ 

.'.  a-^  =  — ,  which  accords  with  the  definition  of  negative  expo- 
nents  (43). 

110,  Cor.  3. — A  factor  may  he  tramfenrdfrom  dividend 
to  divisor  (or  from  numerator  to  denominator  of  a  fraction, 
which  is  the  same  thing),  and  vice  versa,  by  changing  the  sign 
of  its  exponent. 


Thus  TT-  =  -5r. ;  for  —-  =  -  =  --^&^  =  — -        Thus  also   ■=-. 
=  a-^b-^,  since  "tt  =  ^~^  ^  ^ '  ^^^  ^  —  ^~^- 


EXAMPLES 


Ex.  1.  Show  that      ,  ,  —    „  ^ 


-    and   ar-^  =  —  ,  I  have   — r 
a2  x^  x-^ 


Model  Solution.— Since   «-'  =  -    and   ar-"  =  -  ,  I  have 


^2  ^   _a-'  _¥  jf^  _¥     x"  _  W 
T~^  "  f  ~  a'  '  x''~a^^t~  a^y""' 


X'  X' 


«  T  Tl  1  XU      i.      ^^     ^^     ^V  ^^^ 

2.  In  like  manner  show  that  — -  — 


Dcd~^x~^    „ 

r — 5-5— T    from  nega 

the  process, 


3.  Free  -„      -  from  negative  exponents  and  explain 


DIVISION.  67 

111.  Prob.  1. — To  divide  one  monomial  by  another. 

Rule — Divide  the  nuTnerical  coefficient  of  the 
dividerul  by  that  of  the  divisor,  and  to  the  quotient 
annex  the  literal  factors,  affecting  each  with  an 
exponent  equal  to  its  exponent  in  the  dividend 
minus  that  in  the  divisor,  and  suppressing  all  fac- 
tors whose  exponents  thus  become  0. 

The  sign  of  the  quotient  will  be  +  when  dividend 
and  divisor  have  lihe  signs,  and  —  when  they  have 
unlike  signs. 

Demonstration. — The  dividend  being  the  product  of  divisor  and 
quotient,  contains  all  the  factors  of  both ;  hence  the  quotient  con- 
sists of  all  the  factors  which  are  found  in  the  dividend  and  not  in 
the  divisor.  Or,  the  correctness  of  the  rule  appears  from  the  fact 
that  it  is  the  converse  of  the  corresponding  operation  in  multipli- 
cation, so  that  quotient  and  divisor  multiplied  together  produce 
the  dividend.  The  law  for  the  sign  of  the  quotient  is  demonstrated 
in  (106).     Q.  E.  D. 

EXAMPLES. 

Ex.  1.  Divide  ^ah^b  by  'daW. 

Model  Solution. 
Operation.        3aV )  12aV6 

Explanation. — In  the  divisor  there  is  a  factor  3,  hence  there 
must  be  a  factor  4  in  the  quotient,  to  produce  12  in  the  dividend. 
So,  also,  since  there  are  3  factors  of  a  in  the  dividend,  and  2  in  the 
divisor,  there  must  be  1  in  the  quotient  in  order  that  tiie  product 
of  divisor  and  quotient  may  be  the  dividend.  In  like  manner  4 
factors  of  x  in  the  dividend,  and  2  in  the  divisor,  require  2  iu  the 
quotient.  There  being  1  factor  &  in  the  dividend,  and  none  in  the 
divisor,  one  factor  of  h  must  appear  in  the  quotient.  Hence  12a'aj*& 
-^3aV  =  4aa;"^&,  which  quotient  containing  all  the  factors  of  the 
dividend  not  found  in  the  divisor,  will,  when  multiplied  into  the 
divisor,  produce  the  dividend, 


68  FUNDAMENTAL    RULES. 

2.  Divide  IWcY  ^1  ^^^f-  Quot.,  S¥c. 

3.  Divide  Scdx  by  dx.  Quof.,  Sc, 

4.  Divide  lOa^cx^y  by  5a^.  Qtcot,  2acy. 
6.  Divide  ISa^b^x  by  —3a^x.  Quot,  —6ab. 

6.  Divide  —20a^y^  by  —bx^.  Quot.,  4.xyK 

7.  Divide  —^2a^mY  by  '7aW.  Quot,  —6af. 

8.  Divide    -Sla^bx^y  by  9«%.  Quot.,  —9aa^. 

9.  Divide  —  13a:y  by  —ldxy\  Quot,  x, 

10.  Divide  8aV  by  —  S^j^c^.  gwo^f.,   —  1. 

11.  Divide  3a^m  by  3a%.  §wo2^.,  1. 

12.  Divide  —x^y^  by  a^-^y-^,  §mo^.,   — a:^^^ 

13.  Divide  af^  by  ic^  Quot,  x^-^. 

14.  Divide  5a-^^  by  2a&.  ^woif.,  ^. 

15.  Divide  6ah^  by  3«^Ji  Quot,  'Zar^bk 

16.  Divide  17«-^H.t2  by  lla-^-^x^ 

Quot,  ^^i^x-K 

17.  Divide  aPb^c-'  by  «^>V.          C^o^.,  aP-^b^-^c-i'+S). 


112.  Prob.  2. — To  divide  a  polynomial  by  a  monomial. 

Rule. — Divide  each  term  of  the  polynomial  divi- 
dend hy  the  monomial  divisor,  and  write  the  results 
in  connection  with  their  own  signs. 

Demonstration. — This  rule  is  simply  an  application  of  the  corol- 
laries (103,  104) ;  since  to  divide  a  polynomial  is  to  find  the  quo- 
tient of  the  sum  or  difference  of  several  quantities,  which  by  these 
corollaries  is  shown  to  be  the  sum  or  difference  of  the  quotients  of 
the  parts,     q.  e.  d, 


l)iv^isioK.  69 

EXAMPLES. 
Ex.  1.  Diyide  16a^xi-\-24:a^a^b—12a'^xc  by  Sasfi. 

Model  Solution. 
Operation.       ^aa^ )  15aV  +  24aV&  —  12a^xc 
5a'ar-»  +  Sah  —  4a*z-^c 

Explanation. — I  write  the  divisor  on  the  left  of  the  dividend, 
and  the  quotient  underneath  as  a  convenient  form.  Considering 
the  first  two  terms,  the  quotient  of  their  sum  is  the  sum  of  their 
quotients  (103) ;  hence  I  divide  each  separately  and  add  the  re- 
sults, obtaining  5a^x~^  +  Sdb.  Again,  the  quotient  of  the  difference 
between  the  sum  of  these  two  terms  and  the  third  is  the  difference 
of  the  quotients  (104)  ;  hence  I  subtract  from  the  quotient  of  the 
first  two  the  quotient  arising  from  dividing  12a^xc,  which  is  4a*x-'% 
and  have  for  the  entire  quotient  5a^x-^  +  Sab—4:a*x-^c. 

2.  Divide  da^&^—lSa^iti-^6a^  by  Sah. 

Quot,  a^—Qa%  +  2a. 

3.  Divide  Ux'iy^—%7^f—24^y^  by  8a:. 

Quot,  Zxy^—Qfif—Sy\ 

4.  Divide  210^7^— la^x^-^Uax  by  lax. 

QuoL,  3a^x^—ax-\-2. 

6.  Divide  4:2a^-na^+28a  by  Ha. 

QuoL,  6fl2_-J^^.4. 

6.  Divide  m*—24]i:^c-h^Sk^  by  3M 

QuoL,  3k^^^Sk-^c+16k, 

7.  Divide  72a^(^—^Sa^c*—S2a^c^  by  I6a^(^. 

Quot.,  ia?c-^ — Sa^c — 2flt«c». 

8.  Divide  36m*— 48m*  by  4mi. 

Quot.,  dm^-~12mA 

9.  Divide  m^—mJn^  by  m^.  Quot.,  m^—mi^n^\ 
10.  Divide  a^—a^i^-i-a^  by  ai         Quot.,  a^—t^-{-aK 


70  FUNDAMENTAL    RULES. 

11.  Divide  lla^-Sda^  by  ll«i  Quot,  ai-daS\ 

12.  Divide  72mi^— 60mW  by  2imK 

13.  Divide  2aH4«3J  +  2a2^  by  2a2. 

14.  Divide  lla^a^—Sab-j-lSa^  by  3«a;. 

Quot,  i^ax^—--{-6^' 

15.  Divide  a^^+i— a»»+2_aw+3_^m+4  ^y  ^a, 

m  1 

16.  Divide  bx^—lOx-""  -\-15aFy  by  6x\ 


113.  Definition. — A  polynomial  is  said  to  be  arranged 
with  reference  to  a  certain  letter  when  the  term  containing 
the  highest  exponent  of  that  letter  is  placed  first  at  the  left 
or  right,  the  term  containing  the  next  highest  exponent 
next,  etc. 

Illustration. — The  polynomial  6a;y +  4xy*  +  4«^y4-2/*  +  «*,  when 
arranged  according  to  the  descending  powers  of  y,  becomes  y*-\-4:xy^ 
■\-^x^y'^-\-^x^y  +  x*.  In  this  form  it  also  chances  to  be  arranged  with 
reference  to  the  ascending  powers  of  x. 

114.  Prob.  3. — To  pepform  division  when  both  divi- 
dend and  divisor  are  polynomials. 

Rule. — /.  Arrange  dividend  and  divisor  with  ref- 
erence to  the  same  letter. 

II.  Divide  the  first  term  of  the  dividend  hy  the 
first  term  of  the  divisor  for  the  first  term^  of  the  quo- 
tient. 

III.  Subtract  from,  the  dividend  the  product  of  the 
divisor  into  this  term  of  the  quotient,  and  bring 
down  as  many  terms  to  the  remainder  as  may  be 
necessary  to  form  a  new  dividend.  Divide  as  before, 
and  continue  the  process  till  the  worh  is  complete. 


DlVlSION^.  71 

The  demonstration  of  this  rule  will  be  more  readily  com- 
prehended after  the  solution  of  an  example. 

Ex.— Divide  6a^^  +  x^'-^a:^-\-a*—4:a^  by  a^-{-a^—2ax. 
Model  Solution. 

DIVISOR.  DIVIDEND.  QUOTIENT. 

a*  —  ^a'x  +    gV 

—  2a'a;  +  5aV  —  4aa;' 

—  ^a'x  +  4aV  —  2ax^ 

a'x'  —  2ax^  +  x* 
a'x'  —  2ax'  +  a!' 

Explanation. — Having  arranged  the  dividend  and  divisor  with 
reference  to  the  descending  powers  of  a,  and  placed  the  divisor  on  the 
left  of  the  dividend,  I  divide  a\  the  first  term  of  the  arranged  divi- 
dend, by  a",  the  first  term  in  the  arranged  divisor,  and  get  a-  as  the 
highest  power  of  a  in  the  quotient.  Now,  as  I  want  to  find  how  many 
times  a''—2ax  +  x'^  is  contained  in  the  dividend,  and  have  found  it 
contained  a'  times  (and  more),  I  can  take  this  a"  times  the  divisor 
out  of  the  dividend,  and  then  proceed  to  find  how  many  times  the 
divisor  is  contained  in  what  is  left  of  the  dividend.  Hence  I  mul- 
tiply the  divisor  by  a'  and  subtract  it  from  the  dividend,  leaving 
—2a^x-\-^a^x'^—4ax^  +  x*.  The  same  course  of  reasoning  can  be  ap- 
plied to  this  part.  Thus  I  know  that  the  next  highest  jDower  of  a 
in  the  quotient  will  result  from  dividing  the  first  term  of  this  re- 
mainder by  the  first  term  of  the  divisor,  etc.  When  this  pro- 
cess has  terminated  I  have  taken  a',  and  —2ax,  and  a?"  times  the 
divisor  out  of  the  dividend,  and  finding  nothing  remaining,  I  know 
that  the  dividend  contains  the  divisor  just  a''—2axi-x-  times. 

We  will  now  give  the  demonstration  of  the  rule. 

Demonstration. — The  arrangement  of  dividend  and  divisor 
according  to  the  same  letter  enables  us  to  find  the  term  in  the  quo- 
tient containing  the  highest  (or  lowest  if  we  put  the  lowest  power 
of  the  letter  first  in  our  arrangement)  power  of  the  same  letter,  and 
80  on  for  each  succeeding  term. 


72  FUNDAMENTAL    RULES. 

The  other  steps  of  the  process  are  founded  on  the  principle,  that 
the  product  of  the  divisor  into  the  several  parts  of  the  quotient  is 
equal  to  the  dividend.  Now  by  the  operation,  the  product  of  the 
divisor  into  the  Jirst  term  of  the  quotient  is  subtracted  from  the 
dividend ;  then  the  product  of  the  divisor  into  the  8eco?id  term  of 
the  quotient;  and  so  on,  till  the  product  of  the  divisor  into  each 
term  of  the  quotient,  that  is,  the  product  of  the  divisor  into  the 
wlwle  quotient,  is  taken  from  the  dividend.  If  there  is  no  remain- 
der, it  is  evident  that  this  product  is  equal  to  the  dividend.  If 
there  is  a  remainder,  the  product  of  the  divisor  and  quotient  is 
equal  to  the  whole  of  the  dividend  excejyt  the  remainder.  And  this 
remainder  is  not  included  in  the  parts  subtracted  from  the  divi- 
dend, by  operating  according  to  the  rule. 


EXAMPLES. 

1.  Divide  x^—3ax^-\-Sa^x—a^  by  x—a. 

2.  Divide  2y^—l^y'^-\-^%y—lQ  by  ?/— 8. 

Qiiot,  2f—3y-\-2. 

3.  Divide  x^—a^x^-\-2a^x—a'^  by  x^—ax-j-a\ 

QuoL,  x^-j-ax—a^, 

4.  Divide  a*+4:¥  by  a^—2ab-\-2i^. 

Operation. 

a^  _  2ah  +  21' )  a*  +  4h'  {a^  +  2ab  +  2¥ 

a*  —  2a'b  +  2aW 


2a'b- 
2a'b- 

-  2aW- 

+  45* 
+  4«5=' 

2a'l)' 
2a'¥ 

-4a5« 
-4a5« 

+  45* 
+  45* 

6.  Divide  8fl2_26aJ  +  15^  by  4.a—U. 

Quot.f  2a— bh, 

6.  Divide  a^—¥  by  a-h.  Quot.,  a^-^ai^l^. 


DIVISIOK.  73 

7.  Divide  a^—ia^x-^6ah^—4^a^-^x*  by  a^—2ax-{-x^. 

QtioL,  d^—'Zax  +  x\ 

8.  Divide  «3^:c8  by  a-\-x.  QuoL,  a^—ax  +  3^. 

9.  Divide  i83^—76ax^—6'kt^-\-106a^  by  2x—3a. 

Quot,  24x^-'2ax—36a\ 

10.  Divide  2a^-\-a—6  by  2fl— 3.  C^o^v  «  +  2. 

11.  Divide  2:2_^7a;^io  by  a; +2.  Qnot.,  a; +  5. 

12.  Divide  a:3— Sa^^— 46a:— 40  by  a; +  4. 

Quot,  a:2— 9a;— 10. 

13.  Divide  a:8_93;2_,. 27.^—27  by  x—3. 

Qvot.,  a:2_6a;  +  9. 

[Note.— In  the  following  the  pupil  will  need  to  observe  whether 
the  terms  are  properly  arranged  or  not  before  commencing  the 
division.] 

14.  Divide  bx^y-\-y^-{-j(^-\-hxy'^  by  4:xy  +  y^-\-x\ 

QuoL,  x-\-y. 

16.  Divide  Qx^y'^—-^xy^—4:a^y-\-y^-\-x^  by  x—y. 

Quot.,  x^  —  37?y-\-3xy'^—y^, 

16.  Divide  6a?*— 96  by  3a;— 6. 

Quot,  2a;3+4a;2+8a;  +  16. 

17.  Divide  3a^lA-3a^y^-b^-\-a^  by  a^-l^-^3ay^-3a^b. 

Quot,  a^  +  3a^  +  '6al^  +  h\ 

18.  Divide  Q^—y^—i>x^y-\-bxy^-ir\Ox?y'^—\OxY  by  x—y. 

Quot,  (x—yy, 

19.  Divide  aP—li^  by  a—h. 

20.  Divide  m*-\-n*  by  m-{-n. 

21.  Divide  32a;5  +  243  by  2a; +  3, 

22.  Divide  ¥—3y*  by  b—y. 

QuoL,  b^  +  lfiy^by^-\-f-^^' 


7*4:       ,  FU2!fDAMEKTAL    RULES. 

23.  Divide  x^-\-px-^q  by  x-\-a. 


x-\-a 

24.  Divide  aH^*  by  a^-\-db^j'%-Y'b\ 

Quot.,  a^—abV^+l^. 

25.  Divide  |2;3+i2^  +  |:c+f  by  \x-\-l.      Quot,  ic^^j. 

Operation. 

ia;  +  1  )  |««  +  «^  +  fa;  +|  (^M;  j 
^x^  +  x" 

l«  +  f 


26.  Divide  a:*+^  by  a;  +  ?/. 

§wo^.,  x^—x^y-^xy'^—y^+^-' 

27.  Divide  a^wi^ga^Z^^+Z^^n  by  a^+h\ 

Quot,  a^-\-1)\ 

Operation. 

28.  Divide  2a3'*— 6a2«^>/i4.6a»^2/i_2^«  by  «'»— J« 

Quot,  2a^—Wb''-^2L^. 

29.  Divide  ^a^s^j^^iS-fx+J  by  ix  +  d. 

Quot.,  a^—^x-^-l. 

30.  Divide  x^—y^  by  x^—y^-         Quot,  x+x^y^  +  y. 

31.  Divide  «— ^  by  a^—bi. 

Quot,  a^  +  ah^  +  a^b^  +  bl 


DIVISION.  ib 

32.  Divide  x^—x^^^x^  +  6x—2x^  by  a;^— 4a;^  +  2. 

Qtiot.,  x—x^. 

33.  Divide  a27n_3am^n^2c2«  by  a'^— c^ 

5     1  1 

34.  Divide  oi?-\-bx-\ 1--3  by  x-\--' 

X       Xr  X 

Quot,  2:2  +  4 4-^2- 

35.  Divide  x^—-.  by  x — ;•  Quot.,  qi?-\-x-\ f--^* 

X^  X  X        QCr 

36.  Divide  ^--2x^-^4-  +  —  by  2a-^x. 

X  2        a     -^ 

n»      T^•    -J      0^         5X^         39    ,  X  1 

37.  Dmde^4-j^^+j^by-,+  -. 

^     ,     3a:2     13^     39^^ 

38.  Divide    a2^(a_l)a:24-(a— 1)  ic3^(^_l)  a4_a4i   by 
a—x. 

Operation. 

a— aj )  a"  +  (a— !)«' +  (a— !)«'  + (a-l)aj*— 35*  (  a  +  a;  +  aj' +  «M;^ 


oar  +(a- 
ax  — 

-IK 
a;' 

ffa:»  +  (a-l)af« 
ax"-          7? 

aa?»+(<r— l)aj* 
oa;'-          a?* 

39.  Divide    a;(a:— 1)  fl3+(a^  +  22-— 2)«2  +  (3a:2-2:3)  ^_a4 
by  a2a;_^.2a— ar^.  gwo/.,  (a;— l)a-f  ic^. 


16 


tUNDAMEKTAL    RtTLES. 


Scholium — This  process  of  division  is  strictly  analogous  to 
"  Long  Division  "  in  common  arithmetic.  The  arrangement  of  the 
terms  corresponds  to  the  regular  order  of  succession  of  the 
thousands,  hundreds,  tens,  units,  etc.,  while  the  other  processes  are 
precisely  the  same,  in  both. 


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CO.     2 

<l-u 

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QQ 

o 


SYNOPSIS    FOR    REVIEW 
Division. 


ScH.  1.— Relation  of  division  and  multiplication. 

ScH.  2.— General  reason  for  a  quotient. 

Cor.  1.— Multiplying  or  dividing  dividend   and 
divisor. 
j  Cor.  2.— Multiplying  or  dividing  dividend. 

Cor.  3.— Multiplying  or  dividing  divisor. 

Cor.  4.— Quotient  of  sum. 

Cor.  5,— Quotient  of  difference. 
.  Definition.— Cancellation. 


Dividend. 
Divisor. 
Quotient. 
Remainder. 

Laws  of 
Signs. 

Laws  of      1^  pjjj^  ^  Q^^  2.— Negative  exponents. 


DEMONSTRATION. 


I  Cor.  1.— Meaning  of  exponent  0. 
■<  Cor.  2.— Negative  exponents. 
(  Cor.  8.— Transferring  exponents. 


1.  To  divide  one  monomial  by  another,     rttle.  Dem. 

2.  To  divide  a  polynomial  by  a  monomial,  rule.  Deu. 
I  3.  To  divide  one  polynomial  by  another,  rule.  Dem. 
[  Scholium. 


Test  Questions. — How  do  negative  exponents  arise,  and  what  do 
they  signify?  What  is  the  value  of  any  quantity  with  0  for  its  ex- 
ponent ?  Why  ?  How  do  you  divide,  when  dividend  and  divisor 
consist  of  the  same  quantity  affected  by  exponents  ?  Why  ?  Give 
the  General  Rule  (Prob.  3)  and  its  demonstration. 


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FUNDAMENTAL    PROPOSITIONS. 

115,  The  Factors  of  a  number  are  those  numbers 

which  multiplied  together  produce  it. 

A  FacU^r  is,  therefore,  a  Divisor.  A  Factor  is  also  frequently  call- 
ed a  meaxure^  a  ter.n  arising  in  Geometry. 

116,  A  Common  Divisor  is  a  commou  integral  fac- 
tor of  two  or  more  numbers. 

The  Greatest  Common  Divisor  of  two  or  more  num- 
bers is  the  greatest  common  integi*al  factor,  or  the  product 
of  all  the  common  integral  factors. 

Common  Measure  and  Common  Divisor  are  equivalent  terms. 

117,  A.  Common  Multiple  of  two  or  more  numbers 
is  an  integral  number  which  contains  each  of  them  as  a  fac- 
tor, or  which  is  divisible  by  each  of  them. 

The  Least  Common  Multiple  of  two  or  more  numbers 
is  the  least  integral  number  which  is  divisible  by  each  of  them. 

118,  A  Composite  Number  is  one  which  is  com- 
posed of  integral  factors  different  from  itself  and  unity. 

119,  A  Prime  Number  is  one  which  has  no  integral 
factor  other  than  itself  and  unity. 

120,  Numbers  are  said  to  be  Prime  to  each  other  when 
they  have  no  common  integral  factor  other  than  unity. 


78  FACTORING. 

Scholium. — The  above  definitions  and  distinctions  have 
come  into  use  from  considering  Decimal  Numbers.  They  are  only 
applicable  to  literal  numbers  in  an  accommodated  sense.  Thus,  in 
the  general  view  which  the  literal  notation  requires,  all  numbers 
are  composite  in  the  sense  that  they  can  be  factored;  but  as  to 
whether  the  factors  are  greater  or  less  than  unity,  integral  or  frac- 
tional, we  cannot  affirm. 

121,  Prop.  1. — A  monomial  may  he  resolved  into 
literal  factors  hy  separaMng  its  letters  into  any  num- 
ber of  groups,  so  that  the  sum  of  all  the  exponents  of 
each  letter  shall  make  the  exponent  of  that  letter  in 
the  given  monomial. 

Illustration,  ^a^lx^  may  be  resolved  into  ^a  •  ah^  •  ab^  •  x^,  or  5a' 
X  &3  X  a^h^x^  X  a^b^x^  x  aj^'  or  into  any  number  of  factors,  in  a  like 
manner. 

Demonstration.— This  is  a  direct  result  of  the  principle  that 
mcmomials  are  multiplied  by  writing  the  several  letters  in  connec- 
tion, and  affecting  each  with  an  exponent  equal  to  the  sum  of  the 
exponents  of  that  letter  in  the  factors. 

EXAMPLES. 

Ex.  1.  Separate  VZa^a^  into  all  the  possible  factors  with 
positive  integral  exponents. 

Factors,  3,  2,  2,  a,  a,  h,  x,  x,  and  x. 

2.  Separate  16a^x^  into  two  equal  factors. 

Factors,  4:a^x^,  and  4:a^x^. 

3.  Separate  Sx^y^  into  two  factors,  one  of  which  is  4xy, 

Factors,  4:xy,  and  2x^1/^. 

4.  Remove  the  factor  2(ax)^  from  6a^x.     Result,  Sa^xk 

5.  Remove  the  factor  3a^  from  15ac^.    Result,  5  ^/ad^, 

6.  Resolve  m  into  two  factors. 

Result,  m^-m^,  ^/m-  ^m,  m^-rr^,  m^-m^,  rf^-ir^,  etc 


FUNDAMENTAL    PROPOSITIONS.  79 

7.  Resolve  x  into  two  equal  factors.       Into  3.     Into  5. 


122,  Prop.  2. — Any  factor  which  occurs  in  every 
term  of  a  polynomial  can  be  removed  by  dividing 
each  term  of  the  polynomial  by  it. 

Demonstration. -This  is  the  ordinary  problem  of  division  by  a 
iiionomial,  since  divisor  and  quotient  are  the  factors  of  the  divi- 
dend. 

EXAMPLES. 

1.  Factor  3«--3^.  Residt,  3  (a— ^). 

2.  Factor  ax—hx-\-cx.  Residt,  (a—h-\-c)x. 

3.  Factor  5—5^.  Result,  6(1— y). 

4.  Factor  6aY—lSaf.  Residt,  Qaf{ay—^). 

5.  Factor  ^'ix^y—l^xy-\-Wy\ 

Result,  7a:^  ( 6a: — 2  +  xhj). 

6.  Factor  %lcm^x—Q^cm7^.      Result,  ^cmx  {^hn  —  lx^). 

7.  Factor  laH^y—Uax^. 

Factors,  laoi?y  and  d^—Zy\ 

8.  Factor  't%aW7?—'^^a¥x^^r'^Wh^:^, 

Factors,  l%ah^x^  and  6a;— 7J  +  8«. 

9.  Factor  924a4^>V-1078«3Z>4c3_^1232a3JV. 

Factors,  UaW(^  and  66«— 77J  +  88c. 


123*  Prop.  3.— If  two  terms  of  a  trinom^ial  are 
POSITIVE  and  the  thir^d  term  is  twice  the  product  of 
the  square  roots  of  these  two,  and  positive,  the  trino- 
mial is  the  square  of  the  sum  of  these  square  roots. 
If  the  third  term  is  negative,  the  trinomial  is  the 
square  of  the  DIFFERENCE  of  the  two  roots. 

Demonstration. — This  is  a  direct  consequence  of  the  theorems 
thit,  "The  square  of  the  sum  of  two  quantities  is  the  sum  of  their 
squares  j!>^M.«  twice  their  product;"  and  "  The  square  of  their  dif- 
ference is  the  sum  of  their  squares  minm  twice  their  product." 
(94,  95.) 


80  FACTORING. 

EXAMPLES. 

Ex.  1.  What  are  the  factors  of  a^  +  2ab-\-i^? 

Model  Solution. — I  observe  that  the  two  terms  a'  and  ¥  of  this 
trinomial  are  both  positive,  and  that  the  other  term,  2a&,  is  twice 
the  product  of  the  square  roots  of  a^,  and  &*,  and  positive.  Hence 
a^  +  2ab  +  I)''  =  {a  +  if  =  {a  +  l){a-\-I>).  Therefore,  the  factors  are 
a  +  1)  and  a  +  6. 

2.  What  are  the  factors  of  a^—2ax-\-a^  ? 

3.  What  are  the  factors  of  m*-^n^-\-2mH^  ? 

Suggestion. — Here  m*  and  n*  are  both  positive,  and  2m';i'  is  the 
product  of  their  square  roots.    Ans.,  im^-\-ji^)  and  (m^  +  n^). 

4.  What  are  the  factors  of  Ua^—Sa-f-1? 

5.  What  are  the  factors  of  m  -f-  2  Vnm  -{-n? 

Ans.,  (Vm  +  Vn)  and  (Vm-i-Vny 

6.  What  are  the  factors  of  a;^+y^— 2a;%t  ? 

Ans.,  x^—y^  and  x^—y^, 

7.  What  are  the  factors  of  a%-\-ab^-^2aU^  ? 

8.  What  are  the  factors  of  x^-\-2xy—y'^'^ 

Query. — Can  the  last  be  factored  according  to  this  Propo- 
dtion  ?     Why  ? 

9.  If  4«2j  16^2^  and  IQab  are  the  terms  of  a  trinomial, 
what  must  be  their  respective  signs  so  that  the  trinomial 
can  be  factored  ? 

10.  Factor  16a^hn^—SaWm-\-l. 

Factors,  (Umm—l)  and  (4a3^m  — 1). 

11.  Factor  Qa^^^ab-^^i^. 

Factors,  (3a -{-^b)  and  (3«  +  i*). 

1«.  Factor  ^da^b^—y^ab^-}-i¥. 

Factors,  {Hab-lb^)  and  (7«^>~|J3), 


FUNDAMENTAL     PROPOSITIONS.  81 

13.  Factor  J  +  2  + J.     Factors,  g  +  ^)  and  g  +  ^). 

14.  Factor  Tt^:i^  +  i^y^—\^y^- 

Farfor.%  (^x^—:^i/)  and  (ix^—iy*). 

15.  Factor  ^+-^-2. 

1 6.  Factor  a^  +  2a\/x  +  x. 

Factors,  (a  +  Vrr)  and  (a-\-Vx), 

17.  Factor  re— 2*^/5+^2. 

Factors,  (Vx—h)  and  (Va;— d). 

18.  Factor  x—'ZVx'-y-^-y. 

Factors,  {Vx—yy)  and  (a/5— V^). 

19.  Factor  a%—'ilax^/h  +  x^. 

Factors,  (a  ^/h—x)  and  {a  Vh—x), 

20.  Factor  a  +  2  Vfl  +  1. 

Factors,  (V«  +  l)  and  (V«-f-l). 


-i^4.  Prop.  4. — T7ie  difference  betweeti  two  quanti- 
ties is  equal  to  the  product  of  the  sum  and  difference 
of  their  square  roots. 

Demonstration. — This  is  an  immediate  consequence  of  the 
theorem,  that  ''  The  product  of  the  sum  and  difference  of  twoqian- 
tities  is  the  difference  of  their  squares."   Thus  a'— 5'  =  (a +  5)  (a—b). 

Also,  if  we  have  m—n,  in  is  the  square  of  w»,  and  n,  of /i^ ;  .-.  m 

^n  =  (m»  +  7i^  (in»— n^),  etc.  ' 

EXAMPLES. 
Ex.  1.  Factor  Ux^-^y^. 

Factors,  (^x—^y)  and  (^x-\-^y). 


83  FACTORING. 

2.  Factor  a^f—l^x^. 

Factors,  (ay—hx)  and  {ay-\-hx), 

3.  Factor  16«%2_25^^2. 

Factors,  (Aax—^by)  and  (4:ax  +  5hy). 

4.  Factor  x^—y^.  Factors,  (x^—y^)  and  (a^^+y^), 

5.  Factor    x^—y^;    also  a^—y^ ;    also   ic^— y* ;    also   a;~* 
— y-*;   also  4«-«— 9J-4. 

Factors  of  last  three,  (x^—y^)  and  (x^  +  y^) ;  (x~'^—y~^) 
and  (x-^+y-^);  and  (2a~^~db-^)  and  {2a-^-^3b-^). 

6.  Factor  m—n. 

Factors,  {"s/m—Vn)  and  (a/wz  +  V^O- 

7.  Factor  2«— 4J2. 

Factors,  2,  (V«— V^J)  and  (V'«+ a/2J). 

8.  Factor  26a^—Sy^^. 

Factors,  {ox^^  —  VSy"^)  smd  (bx'i -{- VSy^). 


125.  Prop.  5. —  When  one  of  the  factors  of  a  quan- 
tity is  given,  to  find  the  other,  divide  the  given  quan- 
tity by  the  given  factor,  and  the  quotient  will  he  the 
other. 

Demonstration. — This  is  the  ordinary  problem  of  division,  since 
the  divisor  and  quotient  are  the  factors  of  the  dividend.  Any 
problem  in  division  affords  an  example. 

EXAMPLES. 

Ex.  1.  Resolve  2a^—dQa%  into  two  factors  one  of  which 
is  2«2. 

13  19 

2.  Remove  the  factor  — „ ^  from  — , ^^ 

x^     1/3  a:*     'ip 

3.  Remove  the  factor  i-  4-  ^r-  from  a^h'"^ -, —  • 

h       2m  4 

9       12      4 

4.  Remove  the  factor  3a~2-f-2:2;~i  from  ~4+  ~3i;+  ^* 


FUNDAMENTAL    PROPOSITIONS.  83 

126.  Prop.  6.  —  The  difference  between  any  two 
quantities  is  a  divisor  of  the  difference  between  the 
same  powers  of  the  quantities. 

The  SUM  of  two  quantities  is  a  divisor  of  the  differ- 
ence of  the  same  even  powers,  and  the  sum  of  the 
same  odd  powers  of  the  quantities. 

Demonstration. — Let  x  and  y  be  any  two  quantities  and  n  any 
positive  integer.  Flrst^  ^—y  divides  x^ — y".  Second,  if  n  is  eve7i^ 
x  +  y  divides  af—y\     Third^  iin  is  odd,  x-\-y  divides  af +  y". 

riBST. 

x—y)  af'—y^       (  a?"-'  +  a?«-V  +  «"~  V  +  a?"-*^''  +  etc. 
a^—x^-^y 


X'^-^y^  -  x^-*y* 

af^*y* — y» 

Taking  the  first  cast  we  proceed  in  form  with  the  division,  till 
enough  terms  to  determine  the  law  are  found.  We  find  that  each 
remainder  consists  of  two  terms,  the  second  of  which,  — y",  is  the 
second  term  of  the  dividend  constantly  brought  down  unchanged  ; 
and  the  first  contains  x  with  an  exponent  decreasing  by  unity  in 
each  successive  remainder,  and  y  with  an  exponent  increasing  at  the 
same  rate  that  the  exponent  of  x  decreases.  At  this  rate  the 
exponent  of  x  in  the  wth  remainder  becomes  0,  and  that  of  y,  n. 
Hence  the  nth  remainder  is  y—y"  or  0;  and  the  division  is  exact. 

SECOND  AND  THIRD. 

x-\-y)X"  ±y"       (  a^-' —3^'-^y  +  x"-^y^—xn-*y^,  etc. 

— a?*~*y  ±  y" 
— af»-V— ^"V 


x^-y  ±  yn 
flf-V+iC'-'y" 


—x"—y  ±  y* 
—3f*~y—x^-*y* 

nf-y  ±  tr 


84  FACTORIN-G. 

Taking  xi-y  for  a  divisor,  we  observe  that  the  exponent  of  a;  in 
tlie  successive  remainders  decreases,  and  that  of  y  increases  the  same 
as  before.  But  now  we  observe  that  the  first  term  of  the  remainder 
is  —  in  the  odd  remainders,  as  the  1st,  3rd,  5th,  etc.,  and  +  in  the 
even  ones,  as  the  2nd,  4th,  6th,  etc.  Hence  if  n  is  even,  and  the 
second  term  of  the  dividend  is  —y",  the  ?ith.  remainder  is  y'^—y"  or  0, 
and  the  division  is  exact.  Again,  if  n  is  odd,  and  the  second  term 
of  the  dividend  is  +y",  the  nth  remainder  is  —y"  +  y''  or  0,  and  the 
division  is  exact,     q.  e.  d. 

127»  Scholium. — The  pupil  should  notice  carefully  the  form 
of  the  quotient  in  each  of  the  above  cases,  and  be  able  to  write  it 
without  dividing.  He  should  also  be  able  to  tell  at  a  glance, 
whether  such  a  division  is  exact  or  not,  and  if  not  exact,  what  the 
remainder,  or  fractional  term  of  the  quotient  is. 

EXAMPLES. 
Write  the  quotients  in  the  following  without  dividing  : 

1.  {a^  -  «/5)  ^  (x  —  y). 

2.  («3  ^_  58)  ^{aJr  h).  Quot,  a^-ab  +  l^. 

3.  («3  _  j3)  -^  (a-  h). 

4.  {a^  —  m^)  -7-  (a  +  tn). 
6.     (a^  —  771^)  -^  (a  —  m). 

6.  {nf  +  n')  -^  {m  -\-  n). 

7.  {m^  —  ¥)  -=-  (?n  -\-  b),  also  by  (m  —  h). 

8.  (x^  -  ^12)  ^  {x^  4-  2f). 

Suggestion. — Notice  that  x^"^  is  the  4th  (an  even)  power  ofx^,  and 
y^'^  is  the  sa)ne  power  of  y^.  It  may  be  best  for  the  pupil  to  obtain 
this  quotient  thus.  Put  x^  =  a,  and  y^  =  h.  Then  x^^  =  a*  and 
f"  =  l\  But  (a*— &')-h  (a  +  5)  =  a^—a''l)  +  db''-'b\  Whence  restor- 
ing the  values  of  a  and  &,  we  have  {x^'^—y^'^)^{x^-\-y^)  =:x^—xhf 
+x^y^—y^,  since  a^  =  x\  —a^l)  =  —x^y^,  +db'^  =  -\-x^y^,  and  —¥ 
=  -/. 

Quot,  m^ — m^n^  -f  m^n^—m^n^  -\-  n^, 

10.  (a;i8_2/i8)-j-(;7;3_^3)^  also  by  (x^  +  y^). 

1 1 .  (ahn^ — y^)  -f-  (atn^ —y)- 


FUNDAMENTAL    PROPOSITIONS.  85 

Suggestion. — Notice  that  d^m^  is  the  third  power  of  am"^,  as  y^  is 
of  y.  The  quotient  may  be  obtained  by  first  using  a  single  letter 
for  aw",  as  x,  writing  out  {x'^—y^)-^(x—y)  =  x^  +  xy-^-y^,  and  finally 
restoring  the  value  of  x.  After  a  little  practice  the  pupil  will 
not  need  to  go  through  such  a  process,  but  can  write  the  quotient 
at  once.     This  quotient  is  a-m*  -\-aiiry-\-y-. 

12.  (i-if)^(l-y). 

13.  (/_l)-^(y  +  l),  also  by  (y-1). 

14.  {a^7p-^¥Y^)-^(ax  +  Wy^). 

15.  (162:4-81y8)^(2a;-h3y2)^  also  by  (2a;— 3^2). 

[Note. — In  the  following  examples  if  the  division  is  not  exact, 
the  quotient  should  be  written  with  the  fractional  term.] 

16.  Is  a}-{-h'  exactly  divisible  by  «  + J,  or  a-- J? 

17.  Is  m^—y^  exactly  divisible  by  m  +  y,  or  m—y  ? 

18.  Write  the  following  quotients  (x^^—y^^)-^(x-\-y), 
also  hj  x—y.    (x^'^ + ^  -f- y^w +\^^(^x-\-y),  m  being  an  in teger. 

Query. — Is  %m  even  or  odd?    Is  2m  +  l  even  or  odd? 


FOR   REVIEW    OR    ADVANCED    COURSE. 

[Note. — The  following  corollary  and  examples  may  be  omitted 
till  after  the  pupil  has  been  through  the  chapter  on  Powers  and 
Roots,  if  thought  desirable.] 

128.  Cor. — Proposition  6  applies  equally  to  cases  involving 
fractional  or  negative  exponents. 

Thus,  a;^— y^  divides  x^-y^,  since  the  latter  is  the  difference 
between  the  4th  powers  of  «'  and  y*.      So  in  general  x  ^—y  »• 

an  lu 

divides  x  "*— y   •■,  a  being  any  positive  integer.      This  becomes 
evident  by  putting  x  "^  —  %  and  y  •"  =  wj  ;  whence  a?"'*  =  «",  and 

at  _£?  _£f 

y   T  =  w"".     But  v^—W  is  divisible  by  v—w^  hence  x  ♦"— y    "■   is 
divisible  by  a;  *"— y  '. 


FACTORING. 


EXAMPLES. 


19.  What  is  the  quotient  of  (tr^— /;-5)-^(«-l— ^>-i)  ? 

20.  What  is  the  quotient  of  {x~^—y~^)-^{x~^-^y~^)? 

Ans.,  x'^—x'^y"^  +  a;~%"» — «/~5. 

21.  What  is  the  quotient  of  [^-^j-^  [|+^^]? 

22.  Is  x~^—y~'^  divisible  by  x~^—y~^,  ov  x~^ -{- y~'^^ 

23.  Is  x^—y^  divisible  by  Vx-j-Vy,  or  Vx—Vy? 

24.  Is  x^-i-y^  divisible  by  \/x-\-\/y,  or  Vx—Vy  ? 

25.  What    is   the   quotient    of    {ax^  +  i^y)  -^  (a^  \/x 


129,  Prop.  7. — A  trinoTviial  can  he  j^esolved  into 
two  binomial  factors,  when  one  of  its  tenns  is  the 
product  of  the  square  root  of  one  of  the  other  two, 
into  the  sum  of  the  factors  of  the  remaining  term. 
The  two  factors  are  severally  the  algebraic  sum  of  this 
square  root,  and  each  of  the  factors  of  the  third  term. 

Illustration. — Thus,  in  a;^  +  7a;  +  10,  we  notice  that  llx  is  the  pro- 
duct of  the  square  root  of  a?",  and  2-1-5  (the  sum  of  the  factors  of  10). 
Therefore,  the  factors  of  x^ -I- 7a; -f- 10  are  a!-f-3  and  a?-|-5.  Again  a;- 
—3a;— 10,  has  for  its  factors  a!-l-2  and  a;— 5,  — Sa;  being  the  product 
of  the  square  root  of  a;^  (or  a?),  and  the  sum  of  —5  and  2  (or  —3), 
which  are  factors  of  —10.  Still  again,  a;''-f-3a!— 10  —  {x—2)  (a7-i-5), 
determined  in  the  same  manner. 

Demonstration. — The  truth  of  this  proposition  appears  from 
considering  the  product  of  x^a  by  a;-i-&,  which  is  x^ ^- {a -\-'h)x -\- db. 
In  this  product,  considered  as  a  trinomial,  we  notice  that  the  term 
(a  +  5)a!,  is  the  product  of  'y/x'^  and  a-h&,  the  sum  of  the  factors  of 


FUNDAMENTAL    PROPOSITIONS.  87 

ab.  In  like  manner  (x-\-a)  (x—h)  =  ^'^ -\-{a—l)x—ah^  and  (pt—a) 
{x—h)  =  a^—(a+b)x  +  afK  both  of  which  results  correspond  to  the 
enunciation,     q.  e.  d. 

[Note. — In  application,  Proposition  7  requires  the  solution  of  the 
problem :  Given  the  sum  and  product  of  two  numbers  to  find  the  num- 
bers, the  complete  solution  of  which  cannot  be  given  at  this  stage  of 
the  pupil's  progress.  It  will  be  best  for  him  to  rely,  at  present, 
simply  upon  inspection.] 

EXAMPLES. 
Ex.  1.  Show  that,  according  to  this  proposition,  the  fac- 
tors of    x^-\-Sx-\-2    are    x-\-l    and    x-{-2.     Verify  it  by 
actual  multiplication. 

2.  In  like  manner  show  the  following : 

x'i  -  7x  H-  12  =  {x  —  S){x-  4). 
x^-  x  —  U  =  ix  +  3)  (x  —  4). 
x^  -\-    x—12=:{x  —  S)(x-{-  4). 

3.  Factor  xi—5x-{-6,  x^+Sx—^S,  ar^— lla;-f28,  xi-\-2x 
—35,  2^2— 2a:-8,  2:24-20;— 8,  T2—y—y\  5-42—^2. 


130,  Prop.  8. —  We  can  often  detect  a  factor  by 
separating  a  polynomial  into  parts. 

Demonstration. — The  correctness  of  this  process  depends  upon 
the  principle  that  whatever  divides  the  parts,  divides  the  whole. 

EXAMPLES. 

Ex.  1.  Factor  a^+12x—2S. 

Model  Solution —The  form  of  this  polynomial,  suggests  that 
there  may  be  a  binomial  factor  in  it,  or  in  a  pwt  of  it.  Now  «' 
— ix-^i  is  the  square  of  x—2,  and  (a-'— 4a;  +  4j-|-(16«— 32)  makes 
a;-^ 4. 12a;— 28.  But  (x''-4x  +  4)  +  (iex—Z2)  =  {x-2){x—2)  +  (x-2) 
16  =  {x-2)  (aj— 2-1-16)  =  (x—2)  (x-\-U).  Whence «— 2,  and  aj-f  14 
are  seen  to  be  the  factors  of  a;^-|-12a;— 28. 

2.  Resolve  12a'^b-\-dby^—loaby  into  its  factors. 

Solution.  3&  is  seen  to  be  a  common  factor,  and  removing  it,  we 
have  4u'^  +  y'^—6ay.    This  latter  polynomial  may  be  separated  into 


88  PACTOEiNa. 

the  parts  a'-'%ay-\-y'^^  and  3«'— Say.  Hence,  4a^  +  y2— Say  =  {a" 
-2a2/  +  y'^)  +  (3a^-3ay)  =  {a-y)  {a-y)  +  {a-y)M  =  («-y)  {a-y 
4-3a)  =  {a—y){4ta~y).  Hence  the  factors  of  13a-fe  +  3%''— 15<%, 
are  35,  a—y,  and  i.a—y. 

3.  Factor  6««+15fl54^— 4^3^2—1  O^^dc^. 

Factors,  o?,  Sa^—2o^,  and  2a +  5 J. 

4.  Factor  3a2_2«5— 1.  Factors,  a—1  and  Sa  +  l. 
6.  Factor  6a^—Sax-\-3xl    Factors,  5a— 3x  and  a—x. 

6.  Factor  20aa;—25«?^— 165a;  +  20^72. 

7.  Factor  Sd^-^22-ab-i-15bK 

8.  Factor  12:»3_8^^_9^y  _|_6^3 

9.  Factor  ^2x^—2xy—20y\  4:2x^—6Sxy-\-20f,  and  ^2x^ 

10.  Factor  64:xy—10ax  +  96by—16ab. 

11.  Factor  Aa; — a^mn — a^xy-\-a^my. 

12.  Factor  ^5a^3?-{-3Wbx^y—21a:kxhj  —  l^abcxy\ 

13.  Factor  12a;5_8a;3^2^21a;2y— 14^3, 


:^^2.  The  Highest  Divisor  of  a  literal  number  is  the 
divisor  of  tlie  highest  degree  with  reference  to  the  letter  of 
arrangement,  i.  e.,  the  divisor  wliich  involves  the  highest 
exponent  of  that  letter. 

The  Highest  Common  Divispr  of  two  or  more  literal 
numbers  is  the  divisor  of  highest  degree  which  is  common 
to  all  of  them. 


HIGHEST    COMMON    DIVISOR.  89 

Scholium. — While,  with  reference  to  decimal  numbers,  the  term 
greatest  divisor  is  appropriate,  it  is  scarcely  proper  to  apply  it  to 
literal  quantities,  for  the  values  of  the  letters  not  being  fixed, 
or  specific,  great  or  small  cannot  be  affirmed  of  them.  Thus, 
whether  a^  is  greater  than  «,  depends  upon  whether  a  is  greater  or 
less  than  1,  to  say  nothing  of  its  character  as  positive  or  negative. 
So,  also,  we  cannot  with  propriety  call  a^—y^  greater  than  a—y.  If 
a  =  ^,  and  y  —  ^,  a^—y^  =  t^,  and  a—y  —  i ;  .-.  in  this  case  a^ 
—y^<a—y.  Again,  if  a  and  y  are  both  greater  than  1,  but  a<y, 
a^—y^  though  numeiicaUy  greater  than  a—y  is  abaohttely  less,  since 
It  is  a  greater  negative. 

[Note. — The  general  rule  for  finding  the  Greatest  or  Highest 
Common  Divisor  is  founded  upon  the  four  following  lemmas.] 

132.  Lemma  1. — The  Greatest  or  Highest  G.  D.  of  two  or 
more  numbers  is  the  product  of  their  common  prime  factors. 

Demonstration. — Since  a  factor  and  a  divisor  are  the  same  thing, 
all  the  common  factors  are  all  the  common  divisors.  And,  since 
the  product  of  any  number  of  factors  of  a  number  is  a  divisor  of 
that  number,  the  product  of  all  the  common  pnme  factors  of  two  or 
more  numbers  is  a  common  divisor  of  those  numbers.  Moreover, 
this  product  is  the  Greatest  or  Highest  C.  D.,  since  no  other  factor 
can  be  introduced  into  it  without  preventing  its  measuring 
(dividing)  at  least  one  of  the  given  numbers,     q.  e.  d. 

EXAMPLES. 
Ex.  1.  What  is  the  G.  C.  D.  of  48,  108,  and  72  ? 

Model  Solution.  48  =  2  2  2  2  3,  108  ==  2  •  2  •  3  •  3-  3,  and  72 
=  2  •  2  •  2  •  3  •  3.  The  comm(m  factors,  or  divisors,  are  2  •  2  •  3 ;  hence, 
2-  2-  3  =  12,  divides  all  the  numbers,  and  is  a  C.  D.  Moreover,  as 
there  is  no  ottier  common  factor,  if  we  introduce  an  additional 
factor  into  2  2  •  3,  it  will  not  divide  the  given  number  or  numbers 
which  do  not  contain  this  factor.  Thus,  if  we  introduce  a  factor  5, 
it  will  not  divide  any  one  of  the  given  numbers.  If  we  introduce 
another  factor,  2,  and  have  2  2- 3  •  2  =  24,  it  will  not  divide  108, 
which  has  but  two  factors  of  2.  ..  2  •  2  •  3  =  12,  is  the  G.  C.  D.  of 
48,  108,  and  72. 

[Solve  the  following,  giving  the  explanation  as  above.] 

2.  Find  the  G.  C.  D.  of  84,  126,  and  210.   G.  C.  D.,  42. 


90  FACTORING. 

3.  Find  the  G.  C.  D.  of  70,  105,  and  245.    Q,  0.  D.,  35. 

4.  Find  the  Highest  0.  D.  of  lUhh  and  25^V. 

Solution. — Here  we  see  at  once  the  Highest  Common  Divisor 
with  reference  to  the  lettere,  as  &,  5,  and  c  are  all  the  common 
literal  factors.  There  is  no  common  factor  in  12  and  25.  Hence 
¥c  is  the  H.  C.  D.,  inasmuch  as  no  other  factor  can  be  introduced 
into  this  product  (b'^c)  and  it  still  remain  a  divisor  of  12(^)^0  and 
25&V^ 

6.  Find  the  H.  C.  D.  of  na^l^c^  and  Sa^m, 

H.  C.  D.,  ^aWc. 

6.  Find  the  H.  C.  D.  of  Qa^x^,  l^aHy,  and  Ua^a^y\ 

7.  Find  the  H.  C.  D.  of  ^x^y^  and  ^rnnxy. 

H.  C.  D.,  xy^. 

8.  Find  the  H.  C.  D.  of  U^x—Qahx-\-^y^x  and  ^a^y—4J^y. 

Suggestions.  M'^x—Qdbx+W^x  =  dcc{a—l))  {a—h);  and  4a^y—i¥y 
=  4y{a  +  l)){a—i).     .-.  H.  C.  D.  is  a—i. 

9.  Find   the   H.    0.    D.    of    a:2  +  12a:— 28    and    a^  +  9x^ 

+  27a;— 98. 

Suggestions. — Factor  the  polynomial  of  the  lower  order  first,  as 
it  is  more  easily  resolved.  a;'^  +  12a;— 28,  may  be  factored  by  (129) 
or  (130).  According  to  the  latter  process  we  have  a;^  +  i2aj— 28 
=  x''—^x-{-4:  +  {lQx—d2)  =  {x-2f  +  16{x-2)  =  {x  — 2)  (x-2  +  lQ) 
=  {x—2)  {x-\- 14).  Now  find  by  trial  whether  either  or  both  of  these 
factors  are  divisors  of  the  other  polynomial,  x—2  is,  and  a; +  14  is 
not.     .-.  x—2  is  the  H.  C.  D. 

10.  Find  the  H.  C.  D.  of  3a^x^y—3a^xy—36a^y  and 
3«2^3_48«2^_3^2^2_l_48a2.  ff^  a  D.,  3a^x—12aK 

133.  Scholium. — The  difficulty  of  factoring  renders  this  pro- 
cess impracticable  in  many  cases.  There  is  a  more  general  method. 
But,  in  order  to  demonstrate  the  rule,  we  must  prove  three 
additional  lemmas. 

134.  Lemma  2. — A  polynomial  of  the  form  kx^ -{■Bx^-'^ 

+  Cx^-2 Ex  +  F  which  has  no  common  factor  in  every 

term,  has  no  divisor  of  its  own  degree  except  itself 


HIGHEST    COMMON    DIVISOR.  91 

By  this  form  it  is  meant  that  all  like  powei-s  of  the  letter  of 
arrangement  are  collected  into  a  single  term,  and  that  the  j3olyno- 
mial  is  arranged  according  to  some  letter,  as  for  division. 

Demonstration. — 1st.  Such  a  polynomial  cannot  have  one  factor 
of  the  ;/th  degree, — its  own, — with  reference  to  the  letter  of  arrange- 
ment, and  another  which  C(mtains  the  letter  of  arrangement,  for 
the  product  of  two  nuch  factors  would  be  of  a  higher  (or  different) 
degree  from  the  given  polynomial. 

2d.  It  cannot  have  a  factor  of  the  nth  degree  with  reference  to 
the  letter  of  arrangement,  and  another  factor  which  does  not  con- 
tain that  letter,  for  this  last  factor  would  ajipear  as  a  conmion 
factor    in    every    term,    which    is    contrary    to    the    hypothesis. 

Q.  E.  D. 

Illustration.  4«'— 5ay4-y'  cannot  have  a  factor  containing  a", 
such  as  a?—y,  and  another  which  contains  a,  as  4a— 2y,  since  two 
aacli  factors  multiplied  together,  would  give  a  higher  power  of  a 
than  rt^,  to  say  nothing  of  other  terms.  Again,  it  cannot  have  such 
a  factor  as  a^—ay—y  and  another  which  does  not  contain  a,  as 
4—5y-\'y^,  since  this  last  would  then  appear  as  a  factor  in  every 
tenn  of  the  product  when  arranged  with  reference  to  a;  as 
(4— 5y  +  y-)a' — (4— 5y-j-y^)ay— (4— 5y-f  y")y.  But  the  hypothesis 
is  that  the  given  polynomial  shall  have  no  common  factor  in 
every  term. 

13/>,  Lemma  3. — A  divisor  of  any  number  is  a  divisor 
of  any  multiple  of  that  number. 

Illustration. — This  is  an  axiom.  If  a  is  contained  in  h,  q  times, 
it  is  evident  that  it  is  contained  in  n  times  6,  or  nb,  n  times  j,  or 
nq  times. 

130,  Lemma  4. — A  common  divisor  of  two  numbers  is  a 
divisor  of  their  sum  and  also  of  their  difference. 

Demonstration. — Let  a  be  a  C.  D.  of  m  and  n,  contained  in 
m,    p   times,   and  in   n,  q  times.      Then   (m  -\-  n)  -i-  a  =  2^  +  q. 

Q.  E.  D. 

Illustration. — If  this  appears  a  little  abstract  and  unsatisfactory 
to  the  learner,  let  him  illustrate  it  thus :  4  is  contained  in  30, 
5  times,  and  in  8,  2  times;  hence  it  is  contained  in  20  +  8,  5  +  2,  or 
7  times,  and  in  20—8,  5  —  2,  or  3  times. 


92  FACTORIN^G. 

137,  Prob.— To  find  the  H.  C.  D.  of  two  polynomials 
without  the  necessity  of  resolving  them  into  their  prime 
factors. 

Rule. — /.  Arranging  the  polynomials  with  refer- 
ence to  the  same  letter,  and  uniting  into  single  terms 
the  like  powers  of  that  letter,  remove  any  common 
factor  or  factors  which  may  appear  in  all  the  terms 
of  both  polynomials,  reserving  them  as  factors  of  the 
H.  C.  B. 

II.  Reject  from  each  polynomial  all  factors  luhich 
appear  in  it  and  not  in  the  other. 

III.  Talcing  the  polynomials,  thus  reduced,  divide 
the  one  with  the  greatest  exponent  of  the  letter  of 
arrangCDient,  by  the  other,  continuijig  the  division 
till  the  exponent  of  the  letter  of  arrangement  is  less 
in  the  remainder  than  in  the  divisor. 

IV.  Reject  any  factor  ivhich  occurs  in  every  term 
of  this  remainder,  and  divide  the  divisor  by  the 
remainder  a^s  thus  reduced,  treating  the  remainder 
and  last  divisor  as  the  former  polynomials  •  were. 
Continue  this  process  of  rejecting  factors  from  each 
term  of  the  remainder,  and  dividing  the  last  divisor 
by  the  last  remainder  till  nothing  remains. 

If,  at  any  time,  a  fraction  ivould  occur  in  the  quo- 
tient, multiply  the  dividend  by  any  number  that  will 
avoid  the  fraction. 

The  last  divisor  multiplied  by  all  the  first  reserved 
common  factors  of  the  given  polynomials,  will  be  the 
H.  C.  D.  sought. 

Demonstration. — We  will  first  give  a  demonstration  of  this  rule 
by  means  of  a  particular  example.  Let  it  be  required  to  find  the 
H.  C.  D.  with  reference  to  «,  of  the  following  polynomials: 

12a^^>'  +  35V-15a%+13«%  +  35y''-15«&y^   and    QaW—^o'Vy 


HIGHEST    COMMON    DIVISOR.  93 

• 

Operation. 

t2a'b'  +  Wy'  -  ^5ab'y  +  12«%  +  3&y»  -  I5ahy'  (A) 

6a'b'-6a'h-'y-2bY  +  ^^Y  +  6a%- 6a%'-26y«  +  aofty"  (B) 

4a''b  +  Y-  ^aby  +  4a  V  +  y'  -  5ay''  (^) 

Sa^^  -  3a%  -  g^'  +  «&y'  +  3a'y  -  3aV  -  y*  -j-  ay»  (2)) 

(46  +  4y)  a'  -  (5&y  +  Sy'')  a  +  (Jt/^  +  y'')  {E) 

(36  +  3y)  a»  -  {Zby  +  3y^)  a''  +  Qyy^  +  y") «  -  (^y"  +  y*)  (/?') 

(G')  (ff) 

4a'' — 5ya  4-  y'^  )  3rt^  —  3ya'  +     y'^ci  —     y ' 

4 


13a»-12ya''+  4y''a-  4y'(3a 
12a»-15ya^+   3y^« 

3ya'+     y''^-  4y'' 
4 

(Jf) 

12ya'+  4y'a-16yH3y 
12ya''-15y'^a+  3y=' 

{0) 

Reject  19y».               19y^«-19y'        ((?) 

a— y)4«''— 5ya  +  y'(4a— y 
4a»-4y« 

-  ya+y' 

—  ya+y"" 

.:  The  H.  C.  D.  of  (A)  and  (B)  is  (&)(6  +  y)  (a-y)  =  ah'^+atry 
-h'y-by\ 

Reasoning. 

1st.  Removing  h  from  both  {A.)  and  (B),  and  reserving  it  as  a 
fector  of  the  H.  C.  D.  (Lem.  1),  and  rejecting  3  from  (A),  and  % 
from  (B),  since  they  are  not  common  factors,  and  hence  cannot  enter 
into  the  H.  C.  D.,  we  get  (C)  and  (D). 

2nd.  Arranging  (C)  and  (D)  with  reference  to  a,  the  letter  with 
respect  to  which  the  H.  C.  D.  is  sought,  both  for  convenience  in 
dividing,  and  to  observe  if  any  other  common  factor  appears,  we 
have  (E)  and  (F).  In  these  we  readily  discover  and  remove  the 
common  factor  (h  +  y),  reserving  it  as  a  factor  of  the  H.  C.  D.,  and 
get  (G)  and  (H). 

3rd.  Having  now  found  two  of  the  common  factors  of  (A)  and 
(B),  and  removed  some  which  were  not  common,  it  remains  to 
determine  whether  there  are  any  more  common  factors,  that  is, 
whether  (G)  and  (H)  have  a  C.  D. 


94  FACTORING. 

(G)  is  its  own  H.  D.,  Lem.  2  ;  hence  if  it  divides  (H),  it  is  the 
H.  C.  D.  of  (G)  and  (H).  We,  therefore,  try  it.  But,  to  avoid 
fractions,  we  multiply  (H)  by  4,  since,  as  4  is  not  a  factor  of  (G), 
the  H.  C.  D.  of  (G)  and  4  times  (H)  is  the  same  as  of  (G)  and  (H). 
We  thus  obtain  (I).  .-.  If  (G)  divides  (I)  it  is  the  H.  C.  D.  of  (G) 
and  (H).  Trying  it,  we  find  the  remainder  (L).  Now,  any  C.  D. 
of  (G)  and  (I)  is  the  divisor  of  (K)  [a  multiple  of  (G)],  Lem.  3,  and 
of  (L),  Lem.  4. 

We  now  have  to  find  the  H.  C.  D.  of  (G)  and  (L),  upon  which  we 
reason  just  as  upon  (G)  and  (H).  Thus,  as  (G)  is  its  own  only  divisor 
of  the  3nd  degree,  if  it  divides  (L),  or  (M)  [since  (M)  =  4  (L)  and  4 
is  not  a  factor  in  (G)],  it  is  the  H.  C.  D.  of  (G)  and  (H).  Trying, 
we  find  a  remainder  (O).  Now,  any  divisor  of  (G)  and  (M)  is  a 
divisor  of  (N),  Lem.  3,  and  of  (O),  Lem.  4.  The  question  is  there- 
fore reduced  to  finding  the  H.  C.  D.  of  (G)  and  (O).  Upon  these 
we  reason  as  before. 

Rejecting  19y^  since  it  is  not  a  factor  of  (G),  and  hence,  cannot 
enter  into  the  H.  C.  D.,  we  have  (P).  The  H.  C.  D.  of  (G)  and  (P) 
cannot  be  higher  than  (P),  and  as  (P)  is  its  own  only  divisor  of  its 
own  degree,  if  it  divides  (G,)  it  is  the  H.  C.  D.  sought.  It  does. 
.-.  a-y  is  the  H.  C.  D.  of  (G)  and  (tl). 

Finally,  &,  (&  +  y),  and  (a—y)  are  all  the  common  factors  of  (A) 
and  (B),  and  hence,  &  (b^y){a—y)~db'^-\-alyy—¥y—ly^  is  their 
H.  C.  D. 

EXAMPLE. 

Find  the  H.  C.  D.  with  respect  to  x,  of  x^—^a^  +  ^lx^ 
— 20a:  +  4,  and  ^7^—l^x^  +  ^lx—lO. 

Model  Solution. — Calling  the  2nd,  (A)  and  the  1st  (B),  I  have 
the  following : 

{A)  Operation.      {B) 

%x^  —  l^x"  +  21a;  -  10  )  «*  —    ^x^  +  21a!-  —  20a;  +  4 

2 


((7)  2a;*  -  16a;^  +  42a!^  _  40a;  +  8  (a;  -  2 

2a!*  —  12a;^  +  21a;''  -  10a!  

IT  4a,3  +  21a;''  _  30a;  +  8 
—    4a!«  +  24a!^  -  42a;  +  20 


{D)        Reject  -3  -    3a;^  +  12a;  -  12 

{E)  x"  -    4a;  +  4 


HIGHEST    COMMON     DIVISOR.  95 

{E)  (A) 

J53  _  4a.  +  4  )  2aj'  -  12a;'  +  21a;  -  10  (  2«  ,-  4 
2x'  -    8a;'  +    8a; 

-  4a;'  +  13a;  -  10 

-  4a;'  +  16a;  -  16 

(F)        Reject  -3  —    3a;  +    6  (E) 

(O)  X—    2)x^  -  4x  +  4:(x  —  2 

a;'  —  2a; 

—  2a;  4-  4 

—  2a;  4-  4 

Reasoning. 

The  two  given  polynomials  being  arranged  with  reference  to  a;, 
and  no  common,  or  other  factor,  appearing,  I  proceed  at  once  to 
determine  by  successive  divisions  their  H.  C.  D. 

The  H.  C.  D.  cannot  be  higher  than  (A),  the  lower  of  the  two ; 
and,  as  it  is  its  own  only  divisor  of  the  3rd  degree,  if  it  divides  (B), 
it  is  the  H.  C.  D.  But  to  avoid  fractions  I  multiply  (B)  by  2,  and, 
since  2  is  not  a  factor  of  (A),  the  H.  C.  D.  of  (A)  and  (B),  is  the 
H.  C.  D.  of  (A)  and  (C).  Now  if  (A)  divides  (C)  it  is  the  H.  C.  D. 
Trying  it,  I  find  a  remainder",  (D). 

Again  the  H.  C.  D.  of  (A)  and  (C)  is  also  a  divisor  of  (D),  for  (D) 
is  the  difference  between  (C)  and  (a;— 2)  times  (A),  both  of  which 
are  divisible  by  the  H.  C.  D.  of  (A)  and  (C).  The  question  is  now 
reduced  to  finding  the  H.  C.  D.  of  (A)  and  (D).  Upon  these  I  reason 
exactly  as  before  upon  (A)  and  (B). 

Thus,  since  —3  is  a  factor  of  (D),  and  not  of  (A),  it  can  be 
rejected ;  and  the  H.  C.  D.  of  (A)  and  (D)  is  the  H.  C.  D.  of  (A) 
and  (E).  This  cannot  be  higher  than  (E) ;  and  as  (E)  is  its  own 
only  divisor  of  the  2nd  degree,  if  it  divides  (A),  it  is  the  II.  C.  D. 
Trying  it,  I  find  a  remainder,  (F).  Upon  this  remainder,  and  (E), 
I  reason  as  before,  upon  (D)  and  (A). 

Thus,  the  II.  C.  D.  of  (E)  and  (A)  is  a  divisor  of  (E)  and  (F),  since 
(F)  is  the  difference  between  (A)  and  a  multiple  (2a;— 4  times)  of 
(E).  The  question  is  then  reduced  to  finding  the  H.  C.  D.  of  (E) 
and  (F).  Upon  these  I  reason  as  before,  rejecting  —3,  which  is  not 
a  common  factor,  and  hence,  forms  no  part  of  H.  C.  T).  of  (E)  and 
(F),  and  finding  by  trial  that  (G)  is  a  divisor  of  (E).  Therefore,  x—2 
is  the  H.  C.  D.  of  (A)  and  (B). 


96  FACTORIJ^G. 


GENERAL  DEMONSTRATION  OP  THE  RULE  FOR 
FINDING  THE  H.  C.  D. 

Let  A  and  B  represent  any  two  polynomials  whose  H.  C.  D.  is 
sought. 

1st.  Arranging  A  and  B  with  reference  to  the  same  letter,  for 
convenience  in  dividing,  and  also  to  render  common  factors  more 
readily  discernif)le,  if  any  common  factors  appear,  they  can  be 
removed  and  reserved  as  factors  of  the  H.  C.  D.,  since  the  H.  C.  D. 
consists  of  all  the  common  factors  of  A  and  B. 

2nd.  Having  removed  these  common  factors,  call  the  remaining 
factors  C  and  D.  We  are  how  to  ascertain  what  common  factors 
there  are  in  C  and  D,  or  to  find  their  H.  C.  D.  As  this  H.  C.  D. 
consists  of  only  the  conmion  factors,  we  can  reject  from  each  of  the 
polynomials,  C  and  D,  any  factors  which  are  not  common.  Having 
done  this,  call  the  remaining  factors  E  and  F. 

3rd.  Suppose  polynomial  E  to  be  of  lower  degree  with  respect  to 
the  letter  of  arrangement  than  F.  (If  E  and  F  are  of  the  same 
degree,  it  is  immaterial  which  is  made  the  divisor  in  the  subse- 
quent process.)  Now,  as  E  is  its  own  only  divisor  of  its  own 
degree  (Lem.  2),  if  it  divides  F,  it  is  the  H.  C.  D.  of  the  two. 
If,  in  attempting  to  divide  F  by  E  to  ascertain  whether  it  is 
a  divisor,  fractions  arise,  F  can  be  multiplied  by  any  number  not  a 
factor  in  E  (and  E  has  no  monomial  factoi'),  since  the  common  factors 
of  E  and  F  would  not  be  affected  by  the  operation.  Call  such  a 
multiple  of  F,  if  necessary,  F'.  Then  the  H.  C.  D.  of  E  and  F',  is 
the  H.  C.  D.  of  E  and  F.  If,  now,  E  divides  F',  it  is  the  H.  C.  D. 
of  E  and  F.    Trying  it,  suppose  it  goes  Q  times,  with  a  remainder,  R. 

4th.  Any  divisor  of  E  and  F'  is  a  divisor  of  R,  since  F'— QE 
=  R,  and  any  divisor  of  a  number  divides  any  multiple  of  that 
number  (Lem.  3),  and  a  divisor  of  two  numbers  divides  their  differ- 
ence. The  H.  C.  D.  divides  E,  hence  it  divides  QE,  and,  as  it  also 
divides  F',  it  divides  the  difference  between  F'  and  QE,  or  R. 
Therefore,  the  H.  C.  D.  of  E  and  F',  is  also  the  H.  C.  D.  of  E.  and  R. 

5th.  We  now  repeat  the  reasoning  of  the  3rd  and  4th  paragraphs 
concerning  E  and  F,  with  reference  to  E  and  R.  Thus,  R  is  by 
hypothesis  of  lower  degree  than  E ;  hence,  dividing  E  by  it,  reject- 
ing any  factor  not  common  to  both,  or  introducing  any  one  into  E, 
which  may  be  necessary  to  avoid  fractions,  we  ascertain  whether  R 
is  a  divisor  of  E,  • 


HIGHEST    COMMOX    DIVISOR.  9? 

6th.  Proceeding  thus,  till  two  numbers  are  found,  one  of  which 
divides  the  other,  the  last  divisor  is  the  H.  C.  D.  of  E  and  F,  since 
at  every  step  we  show  that  the  H.  C.  D.  is  a  divisor  of  the  two  num- 
bers compared,  and  the  last  divisor  is  its  own  H.  D. 

7th.  Finally,  we  have  thus  found  all  the  common  factors  of  A 
and  B,  the  product  of  which  is  their  H.  C.  D.     q.  e.  d. 

EXAMPLES. 
Ex.1.  Find  tlie  H.  0.  D.   of  Uax—Sa-\-aa:^—7ax^  jind 
16a^x^-\-(jah:^—2Sa^a:^,  and  give  the  reasoning  as  above. 

T/ie  H.  a  D.  is  ax—^a. 

2.  Find  the  H.  C.  D.  of  a^-\-Sa^b  +  2aIri  +  ^  and  6(fi 
+  5^,  giving  the  demonstration.       The  H.  C.  D,  is  a-\-l). 

3.  Find  the  H.  0.  D.  o['^Qifi-\-W—^W—l^a^  and  ^la^lf^ 
-Wl^-l^a^yi.  The  K  C,  D.  is  9«4_9a8. 

4.  Find  the  H.  V.  D.  of  ^xy'^—^f  +  Qx^y  and  ^x^y-{-^a^ 
—^xy\  The  K  C.  D,  is  2x  +  2y, 

5.  Find  the  H.  0.  D.  of  3cs^—10x^-^15x-^S  and  :^—2x^ 
—  ^-^^x^+13x-{'(J.        llie  H.  a  D.  is7^-\-3x^-\-dx-^\. 

6.  Find  the  H.  C.  D.  of  W'a^-la^hT^y  +  2alf^xY—^i^f 
and  ^a^l^y^—2ab^xY—2bh^y^' 

The  H.  (L  D.  is  %ax^—2hxy. 


1:^8.  Prob.— To  find  the  H.  C.  D.  of  three  op  more 
polynomials. 

Rule. — Find  the  H.  C.  T>.  of  any  two  of  the  given 
polynomials  hy  one  of  the  foregoing  methods,  and, 
then  find  the  H.  C.  D.  of  this  H.  C.  D.  and  one  of  the 
remaining  polynom^ials.  Continue  this  process  till  all 
the  polynomials  have  been  used. 

Demonstration. — For  brevity,  call  the  several  polynomials  A, 
B,  C,  D,  etc.  Let  the  H.  C.  D.  of  A  and  B  be  represented  by  P, 
whence  P  contains  aU  the  factors  common  to  A  and  B.  Finding 
the  n.  C.  D.  of  P  and  C,  let  it  be  called  P'     P',  therefore,  contains 


98  FACTORING. 

all  the  common  factors  of  P  and  C  ;  and  as  P  contains  all  that  are 
common  to  A  and  B,  P'  contains  all  that  are  common  to  A,  B  and 
C.  In  like  manner  if  P"  is  the  H.  C.  D.  of  P  and  D,  it  contains  all 
the  common  factors  of  A,  B,  C,  and  D,  etc.     q.  e.  d. 

EXAMPLES. 

Ex.  1.  Find  the  H.  C.  D.  of  ''Za:^  +  Q:^-\-^^,  3.t3+9:cH9^ 
+  6,  and  d:^-^Sx^  +  Qx  +  ^.  The  H.  C\  D.  is  x  +  2. 

2.  What  is  the  H.  0.  D.  of  10a^+10aW  +  20a%  2a^-\-2b% 
and  i¥  -j- 12^2^  +  4:a^  +  12ab^  ?  Atis.,  2{a  +  h). 


139 •  The  Lowest  Multiple  of  a  given  literal  number 
is  the  number  of  lowest  degree  with  reference  to  some 
specified  letter^  which  contains  the  given  number  as  a 
factor. 

The  Lowest  Common  Multiple  of  several  numbers  is 
the  number  of  lowest  degree  which  is  a  multiple  of  each. 

[For  the  impropriety  of  the  term  Least  Common  Multiple  as 
applied  to  literal  numbers,  see  Art.  131.J 


140,  Prob.— To  find  the  L  C.  M.  of  two  op  mope 
numbeps. 

Rule. — /.  Tahe  the  literal  numher  of  the  highest 
degree,  or  the  largest  decimal  number,  and  multiply 
it  by  all  the  factors  found  in  the  next  lower  which 
are  not  in  it. 

II.  Again,  iriultiply  this  product  by  all  the  factors 
found  in  the  next  lower  number  and  not  in  it,  and  so 


LOWEST    COMMO^r     MULTIPLE.  99 

continue,  till  all  the  niutibers  are  used.     TJie  product 
tlius  obtained  is  the  L.  C.  M. 

Demonstration. — Let  A,  B,  C,  D,  etc,  represent  any  numbers 
arranged  in  the  order  of  their  degrees,  or  values.  Now,  as  A 
is  its  own  L.  M.,  tlie  L.  C.  M.  of  all  the  numbers  must  contain  it  as 
a  factor.  But,  in  order  to  contain  B,  the  L.  C.  M.  must  contain  all 
the  factors  of  B.  Hence,  if  there  are  any  factors  in  B  which  are 
not  found  in  A,  these  must  be  introduced.  So,  also,  if  C  contains 
factors  not  found  in  A  and  B,  they  must  be  introduced,  in  order 
that  the  product  may  contain  C,  etc.  Now  it  is  evident  that 
the  product  so  obtained,  is  the  L.  C.  M.  of  the  several  numbers, 
since  it  contains  all  the  factors  of  any  one  of  them,  and  hence  can  be 
divided  by  any  one  of  them,  and  ifany  factor  were  removed  it  would 
cease  to  be  a  multiple  of  some  one  or  more  of  the  numbers,    q.  e.  d. 

Ex.  1.  Find  the  L.  C.  M.  of  {x'—l),  (x^—l),  and  (a;  +  l). 

Model  Solution. — The  L.  C.  M.  must  contain  x^—\,  and  as  it  is 
its  own  L.  M.,  if  it  contains  all  the  factors  of  the  other  two,  it  is  the 
required  L.  C.  M.  The  factors  of  x^—\  are  {x—V)  and  («*  +  «+ 1). 
But  the  product  of  these  does  not  contain  the  factors  of  {x^—1)  which 
are  (a^  +  l)  and  («—!).  Hence  we  must  introduce  the  factor  («  +  !), 
giving  {x^—V}  («+l),  as  the  L.  C.  M.  of  a?^ — 1  anda;'*— 1.  Now  as  this 
product  contains  the  third  quantity,  it  is  the  L.  C.  M.  of  the  three. 

2.  Find  the  L.  C.  M.  of  {a^lf,  a^-V",  (a-Vf,  and  «» 

Suggestions. — The  last  is  (a  +  J)^  which  contains  the  factors  of 
the  1st,  but  neither  of  the  factora  of  the  3rd.  Both  lactors  of 
(«—&)'  must,  therefore,  be  introduced,  giving  {a^-Vfia—'by  as  the 
L.  C.  M.  of  the  1st,  3rd  and  4th.  And  as  it  contains  both  factors  of 
«--&^  viz.:  (a  +  &)  (a— &),  it  is  the  L.  C.  M.  of  all. 

3.  Find  the  L.  C.  M,  of  {x^—^a%  (a:-f2a)3,  and  (x—%d)\ 

Z.  C.  if.,  (a:2_4a2)3. 

4.  Find  the  L.  C.  M.  of  Qc^-i^%xy^y^  and  o^—xy\ 

L.  C.  M.,  (x^y)  {a^-xy% 

5.  Find  the  L.  C.  M.  of  l^^m:(^,  S5a^m%  and  7b*x^. 

L.  a  M.,  70a^bWx^, 


100  FACTORING. 

6.  Find  the  L.  0.  M.  of  Uxy,  10a%  and  ^Oa^x\ 

L.  G.  if.,  80a%y. 
Scholium. — In  applying  this  rule,  if  the  common  factors  of  the 
two  numbers  are  not  readily  disceraed,  apply  the  method  of  find- 
ing the  H.  C.  D.,  in  order  to  discover  them. 

7.  Find  the  L.  0.  M.  oi  a^—'Zax^-^-^a^x—Wy  x^-]-^aa'^ 
-\-4:a^x-\-%o^,  and  x^—^a\ 

Model  Solution. — The  L.  C.  M.  of  these  numbers  must  contain 
x^—^ax^  +  Wx—^a^-^  and  as  it  is  its  own  L.  M.,  if  it  contains  all 
the  factors  o^  x^-\-2ax'' +  ^a'x  +  9,a\  it  is  the  L.  C.  M.  of  these  two 
polynomials.  But  as  the  common  factors  of  these  numbers,  if  they 
have  any,  are  not  readily  discerned,  we  apply  the  method  of  H.  C.  D. 
and  find  that  a;^  +  4a^  is  the  H.  C.  D.  of  the  two.  Since,  then,  x' 
— 2aa;'  +  4a'^a;— 8a^  contains  the  factor  a;^  +  4a^  of  the  second  number, 
it  is  only  necessary  to  introduce  the  other  factor  in  order  to  have 
the  L.  C.  M.  of  the  two.  Now  {x"" +  %ax''-\-4:(i'x^Sa^)^{x'' -\-^a') 
=  «  +  2«.  B.QY\CQ  {x^'—'iax'' ■\-ia^x—^a^)  {x-\-%a)  or  a;*— 16a*  is  the 
L.  C.  M.  of  the  first  two  numbers,  since  it  contains  all  the  factors 
of  each,  and  no  more.  Now,  to  find  whether  a;*— 16«*  is  a  multiple 
of  the  remaining  number,  a;"— 4«*,  or,  if  it  is  not,  what  factors  must 
be  introduced  to  make  it  so,  we  proceed  in  the  same  way  as  with  the 
first  two  numbers.  But  our  first  step  (or  124)  shows  us  that  a;* 
—  16a*  is  a  multiple  of  x''—id\  .'.  «*  — 16a*  is  the  L.  C.  M.  of  the 
three  given  numbers, 

8.  Find  the  L.  C.  M.  of  Qx^—x—1  and  '^x^+^x—%. 

L.  a  M„   (2a;24-3a;-2){3a;  +  l). 

9.  Find  the  L.  0.  M.  of  x?—^x^-\-^'^x—lb  and  x?—dx 
+  7. 

L.  C.  M.,  (a;3_9^2_|.232--15)  {x—1)  =  x^-16a^+S6x^ 
~176a;  +  105. 

10.  Find  the  L.  C.  M.  of  2a;-l,  4rr2-l,  and  4a;2  +  l. 

11.  Find  the  L.  0.  M.  of  a^-x,  a^-1,  and  a^-{-l. 

L.  a  M.,  x{a^-l)  =  x'^-x. 

12.  Find  the  L.  C.  M.  of  x^-exi-]-Ux—6,  x^-9xi-h2(jx 
—24,  and  ar^— 82^^-f  19^;— 12. 

L.  0,  M.,  {x-1)  {x-2)  (x-^)  (a;-4)  =  0^-10:^3^35^2 
—50^  +  24. 


BYN0P8TS. 


101. 


SYNOPSIS    FOR    REyiEW. 


Z 

cc 
o 

I- 
o 

< 


to 


oq. 
zo 

LL.a. 


r  Factor.      1  common. 
Divisor.      \  (|.  C.  »[  distinction. 
Measure. 


u.  c.  D. 

COMMON. 


Multiple.     ]  LEAST  C.  M.     Idistdiction 
(  LOWEST  C.  M.  I 

1 


Numbers. 
Scholium 

L  PROPS. 


COMPOSITE. 

PRIME. 

PRIME  TO  EACH  OTHER. 


HOW  THESE  TERMS  ARE  APPLIED. 

1.  A  MONOMIAL.    Dem. 

2.  A  POLYNOMIAL  WITH  MON.  FACTORS.    Dem. 

3.  a»±2a6  +  ^»^    Dem. 

4.  a»-6^    Dem. 

5.  ONE  FACTOR  GIVEN.    Dejt. 

iSc/i.  Form  of  quotient. 
Cor.  Fractional  and  neg- 
ative exponents. 

7.  A  TRINOMIAL.    WHEN  ?    HOW '(    Dem. 

8.  SEPARATING  INTO  PARTS. 


r  Distinction  between  G.  C.  D.  and  H.  C.  D. 
Lemmas. 


1.  DEM. 

2.  DEM. 

3.  DEM. 

4.  DEM. 


General    j  PROB.  l.    Rule.    Dem. 
Method.     1  PROB.  2.    Rule.    Dem. 

I  Distinction  between  Least  and  Lowest  C.  M. 
Problem,      ritle.   dbm. 
Scholium,     by  means  of  h.  c.  d. 


Test  Questions.— What  are  the  factors  of  a^—b'"i  Ofa'  +  2a5 
+  ft-^?  Of  «'-2a5+6''?  or  l-2x  +  x"i  Of  3aH Ba'^a;  +  3a V  ? 
Orx''  +  y--\-2xy'i  Of  a;^— a!-12?  Of  7n?  Of  a'&'y?  Ofa;'-y»? 
State  the  general  rule  for  testing  the  divisibility  of  the  sum  or 
diflfereuce  of  like  powers.  Prove  one  of  the  cases,  as  (a"*—l'")^(a 
—b).  Distinction  between  H.  C.  D.  and  G.  C.  D.  Beivfeen  Lowest 
and  Least  C.  M.  Explain  the  process  of  finding  each  by  factoring. 
Give  the  General  Rule  in  each  case,  and  its  demonstration. 


FRACTIOlfS 


i 


-gfejAFTlII  III. 


* 


ECTiON  I 


DEFINITIONS  AND  FUNDAMENTAL  PRINCIPLES. 

1J,1,  A  Fraction,  in  the  literal  notation,  is  to  be  con- 
sidered as  an  indicated  operation  in  Division. 

A  fraction  is  written,  as  in  common  arithmetic,  with  one 
number  above  another  and  a  line  between  them.  The 
number  above  the  line,  i.  e.,  the  dividend,  is  called  the 
Numeratoi' ;  and  the  number  below  the  line,  i.  e.,  the 
divisor,  is  called  the  Denominator. 


Thus 


3c  +  4a;3 


—  means  nothing  more  than  (3a— 5ma;^)-^(3c+4a;^). 


Taken  together  numerator  and  denominator  are  called 
the  Terms  of  the  Fraction. 

142.  Scholium. — In  the  literal  notation  it  becomes  imprac- 
ticable to  consider  the  denominator  as  indicating  the  number  of 
equal  parts  into  which  unity  is  divided,  and  the  numerator  as 
indicating  the  number  of  those  pai'ts  represented  by  the  fraction, 
since  the  very  genius  of  this  notation  requires  that  the  letters  be  not 

restricted  in  their  signification.     Thus  in  -,  it  will  not  do  to  say,  h 

represents  the  number  of  equal  parts  into  which  unity  is  divided, 
since  the  notation  requires  that  whatever  conception  we  take  of  these 
quantities  should  be  suflSciently  comprehensive  to  include  all  values. 
Hence  &  may  be  a  mixed  number.  Now  suppose  &  =4f.  It  is 
absurd  to  speak  of  unity  as  divided  into  4|  equal  parts. 


DEFINITIONS.  103 

14ii,  The  Value  of  a  Fraction  is  the  quotient  of  the 

numerator  divided  by  the  denominator. 

144,  Cor.  1. — Since  numerator  is  dividend  and  denom- 
inator divisor,  loe  have  (100,  101,  103)  the  following  : 

1.  Dividing  or  multiplying  both  terms  of  a  fraction  by  the 
same  quantity  does  not  alter  its  value. 

2.  Multiplying  or  dividing  the  numerator  multiplies  oi 
divides  the  value  of  the  fraction  ;  and 

3.  Multiplying  or  dividing  the  denominator  divides  or  mul- 
tiplies the  fraction. 

145,  Cor.  2. — A  fraction  is  multiplied  by  its  denominator 
by  simply  removing  the  denominator. 

This  is  simply  cancelling  equal    factors   from  numerator  and 

3  3x4 

denominator  (Art.  105).     Thus  -  x  4  =  — - —  ~  3.    In  like  manner 

4  4 

a  ax 

-Y.x  —  ^  =  a. 

X  X 

14(u  The  terms  Integer  or  Entire  Number,  Mixed  Num- 
ber, Proper  and  Improper,  are  applied  to  literal  numbers, 
but  not  with  strict  propriety.  Thus,  whether  m-\-n  is  an 
integer,  a  mixed  number,  or  a  fraction,  depends  upon  the 
values  of  m  and  n,  which  the  genius  of  the  literal  notation 
requires  to  be  understood  as  perfectly  general,  until  some 
restriction  is  imposed. 

For  convenience,  we  adopt  the  following  definitions : 

147.  A  number  not  having  the  fractional  form  is  said 

to    have    the    Integral    Form;     as    m-f-n,    2o^dSa~^x 
+  3x5^4. 

148.  A  polynomial  having  part  of  its  terms  in  the 
fractional  and  part  in  the  integral  form,  is  called  a  Mixed 

,WT      1                         .  ccP-2y^, 
Number;  as  a—x-\ /—' 


104  FRACTIONS. 

149,  A  Proper  Fraction,  in  the  literal  notation,  is 
an  expression  wholly  in  the  fractional  form,  and  which 
cannot  be  expressed  in  the  integral  form  without  negative 
exponents. 

By  calling  such  an  expression  a  proper  fraction,  we  do  not  assert 
anything  with  reference  to  its  value  as  compared  with  unity.    Thus 

-  is  a  proper  fraction,  though  it  may  be  greater  or  less  than  unity. 

It  may  also  be  written  ab-\ 

150,  An  Improper  Fraction  is  an  expression  in 
the  fractional  form,  but  which  can  be  expressed  in  the 
integral  or  mixed  form  without  the  use  of  negative 
exponents. 

Inus, -——  =E  2a — ^ ;  the  former  of  which  is  called  an 

71— dab  n—dab 

improper  fraction. 

151.  A  Simple  Fraction  is  a  single  fraction  with 
both  terms  in  the  integral  form. 

_^       2x—m^  ^    2cmx^  .      ,    ^     ^. 

Inus  -. -,  and  — --  are  simple  fractions. 

15 2.  A  Compound  Fraction  is  two  or  more  fractions 
connected  by  the  word  of;  but  the  expression  is  not  gener- 
ally applicable  in  the  literal  notation. 

Thus  we  may  write  o  ^f  7  ^'ith  propriety,  but  not  ^  of  — ,  unless 

a  and  h  are  integral,  so  that  the  fraction  y  may  be  considered  as 

2 
representing  equal  parts  of  unity,  as  -  does.     If  the  word  of  is  con- 

3 

sidered  as  simply  an  equivalent  for   x ,  the  notation  is,  of  course, 
always  admissible.    But  it  is  scarcely  a  simple  equivalent. 


FUNDAMENTAL    PRINCIPLES.  105 


153.  A  Complex  Fraction   is  a  fraction  having  in 
one  or  both  its  terms  an  expression  of  the  fractional  form. 

c  ma^ 

^"^'d  ~cd 

Thus  ,  and  kj^  ^J-re  complex  fractions. 

5 — a 

y 

154.  A  fraction  is  in  its  Lowest  Terms  when  there  is  no 
common  integral   factor  in  both  its  terms. 

155.  The   Lowest   Common  Denominator   is  the 

number  of  lowest  degree,  which  can  form  the  denomi- 
nator of  several  given  fractions,  giving  equivalent  fractions 
of  the  same  values  respectively,  while  the  numerators  retain 
the  integral  form. 

156.  Reduction,  in  mathematics,  is  changing  the  form 
of  an  expression  without  changing  its  value. 


SIGNS  OF  A  FRACTION. 


157.  In  considering  the  signs  of  a  fraction,  we  have  to 
notice  three  things,  viz.:  the  sign  of  the  numerator,  the  sign 
of  the  denominator,  and  the  sign  before  the  fraction  as  a 
whole.  This  latter  sign  does  not  belong  to  either  the 
numerator  or  denominator  separately,  but  to  the  whole  ex- 

^,          .       ,,                   .             4:a—6cd      .      ,. 
pression.     Thus,    m    the   expression  — XT'    ^^   "^® 

numerator  the  sign  of  4fl  is  +,  and  of  5cd  — .  In  the 
denominator,  the  sign  of  2x  is  -f ,  and  of  4^^  _|_  j^jgQ^  The 
sign  of  the  fraction  is,  — .  These  are  the  signs  of  operation. 
(56,  33,  34.) 

158.  The  essential  character  of  a  fraction,  as  positive 
or  negative^  can  be  determined  only  when  the  essential 
character  of  all  the  numbers  entering  into  it  is  known. 
It  may  then  be  determined  by  principles  already  given. 
(80, 106.) 


106  FRACTIONS. 


EXAMPLES. 

Ex.  1.  Is  the  fraction  —  o.^  ,  .t  essentially  positive,  or 

negative,  when  a,  m,  x,  and  y  are  each  negative  ? 

Model  Solution.— I.  The  square  of  —a  is  +,  since  it  is  the  pro- 
duct of  an  even  number  of  negative  factors  (S8).    .  •.  4a'^  is  positive. 

For  the  same  reason  mx  is  +•  .'.  3w2a?  is  in  itself  positive,  but 
it  is  made  negative  by  the  —  sign  before  it. 

2.  The  cube  of  —x  is  — ,  since  it  is  the  product  of  an  odd  number 
of  negative  factors  (§J)).     .•.  '^x^  is  negative. 

The  square  of  —y  is  +  (88).     .-.  4«/^  is  positive. 

3.  Finally,  the  essential  sign  of  the  fraction  depends  upon  the 
relative  values  of  a,  ?w,  x,  and  y.  If  4a''' >  3ma;,  and  2x'^>4:y'^^  the 
numerator  and  denominator  have  unlike  signs,  whence  the  value  of 
the  fraction  (143)  would  be  —  ;  but  it  is  made  +  by  the  —  sign 
before  the  fraction. 

Similarly,  if  ia^  <  Zmx,  and  2x^>4:y'^,  the  signs  of  the  numerator  and 
denominator  are  alike,  whence  the  sign  of  the  fraction  would  be  +, 
but  the  —  sign  before  the  fraction  would  change  it  to  —  ;  and  the 
essential  character  of  the  fraction  would  be  negative. 

2.  What  is  the  essential  sign  of       '         ^  ,  when  «=  —  3, 

X  =  —2,  p  =  —4,  and  b  z=  —o?  Ans.,  —. 

3.  What  is   the  essential  siern  of v^ —      when 

°  3a^x 

X  =  —2,  a  =  —1,  and  y  =  3?  Ans.,   — . 

4.  What  is  the  essential  siffn  of ^^-^7 — t. —  ,  when 

®  —2y^—3cx 

X  =  4:,  a  =  —2,  y  —  —1,  and  c  =  Q?  Ans.,   +. 


IfjO.  There  are  five  principal  reductions  required  in 
operating  with  fractions,  viz. ;  To  Loivest  Terms, — From 
Improper  Fractions  to  Integral  or  Mixed  Forms, — From  In- 
tegral or  Mixed  Forms  to  Improper  Fractions, — To  Forms 
having  a  Common  Denominator, — and  from  the  Complex  to 
the  Simple  Form. 

Prob.  1. — To  reduce  a  fraction  to  its  lowest  terms. 
Rule. — Reject  all  cojnrnon  factors  from  both  terms ; 
or  divide  both  terms  by  their  H.  C.  D. 

Demonstration.  — Since  the  numerator  is  the  dividend  and  the 
denominator  the  divisor,  rejecting  the  f^ame  factors  from  each  does 
not  alter  the  value  of  the  fraction  (100).  Having  rejected  all  the 
common  factors,  or,  what  is  the  same  thing,  the  H.  C.  D.  (which 
contains  all  the  common  factors),  the  fraction  is  in  its  lowest 
terms  (154). 

EXAMPLES. 

g^3 Qax^ 

Ex.  1.  Reduce  -^^-^^^^-^  to  ite  lowest  termB. 

Model  Solution. — Resolving  the  terms  of  the  fraction  into  their 

pnrae  factors,  I  have  — -r — — ^ — .r~,-;j  =  ^  ^  ^- tt^-^-    Now, 

cancelling  the  common  factors,  3,  a,  and  a  +  x,  which  is  dividing 
dividend  and  divisor  by  the  same  quantity  and  hence  does  not 

alter  the  value  of  the  fraction,  I  have  — ,  or  -— .     Since  in 

a{a-\-x)         a^  +  ax 

this  there  is  no  factor  common  to  numerator  and  denominator,  it  is 

in  its  lowest  terms  (154). 

op,         15b^y       110m.TV      I2a^x^  irya^r^x^ 

their  lowest  terms. 


108  .FKACTTONS. 

^    ^  ^         15¥xy      llOinx^if     Ua^x^         ^    4:ba^c^x^  , 

2.  Keduce  ^^    /„,   ^^^   .  %,   ^^  o  .,  and    ^^  „  „      to 

their  lowest  terms. 

1  iX!'  36^ 

iJ«.»?<.  (not  given  in  order),  3^,  — ,,  3a^a:,and  -^^. 

3.  Reduce  -,— „  ^-j^^,  and  g^.^^^;^— 9.  to  their 

ic     2  1 

lowest  terms.       Results  (not  in  order),  t-o,  ?7,  and 

^  h^    3  a  +  x 

4.  Reduce    -^^^-,  Ub  +  W  ^  "^^  .tH^^"^"^^  *" 

. ,    .    -         .  ,              T»      1.     x^—hx  n—\       ,  3a2_3aJ 
their  lowest  terms.    Results,  — — r-,  — -^,  and rr 

5.  Reduce   »       »—  to  its  lowest  terms. 

6.  Reduce  TnrrT — ^^rV^  ^^  ^^^  lowest  terms. 

3a;2  +  ^xy  +  3«/2 

7.  Reduce  — ^„ — ^r— -  to  its  lowest  terms. 

^      .,    5(^-1) 
Result,  z~, — —A* 
'  7  (w  +  1) 

8.  Reduce     »  ^^ ; — ^  to  its  lowest  terms. 

y^—%xy-\-a? 

Result,  t±m±^. 

y—x 

9.  Reduce  -s — : 7  to  its  lowest  terms. 

n^—An-\-A: 

rS ^6 

10.  Reduce   -j ^  to  its  lowest  terms. 

Result,  s— -^ 

Scholium  I.— Whenever  the  common  factors  of  the  term  are  not 
readily  discernible,  use  the  process  for  finding  their  H.  C.  D.  (13t). 

11.  Reduce  ^—„  ^-^ ,  ^J,  and  ^-^^^  to  their 
lowest  terms. 

?,  a2  4.«J_|_^,2   ^JT^     __:r_^jr^    and  -2T-:72* 


REDUCTION.  109 


,«    T.  J         6ar*  —  7a;  —  20      Sax^  —  Wax  -\-  3a  , 

12.  Reduce  -r-^ -r^r — -- r ,  ^-^-5 ^-9 577-, ,  and 


242^4-220:2  +  5    ^    ^,    .    , 

to  their  lowest  terms. 


48a:*H-16a;2_15 

Results,  -:   „  ,  »  ,  iTT— r-T —  ?  i"ld 


4a?'+3'  10«  +  5«rc  2a?J-f5a;— 1 

Scholium  2. — Tlie  opposite  process  is  sometimes  serviceable, 

viz. :  the  introduction  of  a  factor  into  both  terms  of  a  fraction  which 

will  give  it  a  more  convenient  form. 

x-4-1  x^ 1 

13.  What  factor  will  change    „         -    to  -3—:;  ? 

Ans.j  a;— 1. 

Ans.,  a-i-b. 

ih'dt  factor  will  change  -^ ^— ^ — 

fl8  +  8rc3 


14.   What  factor  will  change  7 —  to  -^ — r,  ? 


15.  What  factor  will  change  ^-^^^^+^—  to 


16a:*' 


Ans.f  a-\-2x. 


16.  What  factor  will  change  ^ — ^r-^- — jtk to 

°  a^—2ab-\-4:if^ 

-16^^P 


fl3  +  8^  • 

[Note. — It  requires  no  special  ingenuity  to  solve  such  problems, 
since,  if  the  factor  does  not  readily  appear,  it  can  be  found  by  divid- 
ing a  term  of  one  fraction  by  the  corresponding  term  of  the  other.] 


160.  Prob.  2. — To  reduce  a  fpaction  from  an  improper 
to  an  integral  or  mixed  form. 

Rule.—  Perform  the  division  indicated.  (141.) 

Demonstration- — The  operation  is  explained  in  the  same  manner 
as  the  corresponding  case  in  division. 

EXAMPLES. 
Ex.  1.  Reduce  9"~~~  *^  ^^  integral  or  mixed  form. 


110  FRACTIONS. 

Model  Solution. — This  being  an  indicated  operation  in  Division, 
I  have  but  to  perform  the  division.  Now,  since  the  sum  of  the 
quotients  is  equal  to  the  quotient  of  the  sum,  I  have  but  to  divide 
each  term  of  lOax  +  c—h  by  2x,  or  divide  such  as  I  can  and  indicate 
the  division  of  the  others,  and  add  the  results  (112).     Thus  I  find 

that =  5a  +  — ~ . 

2x  2x 

2.  Reduce   — — ! ,    — ^—^ ! — ,   and 

to  integral  or  mixed  lorms. 

r,      T,     n       4:a — 6"       ^     ah        ,         „       ^      ^ 

Results,  3a — ,  6y-\-l-\-—  ,  and  a— 3c— 2  +— • 

oa  X  oa 

3.  Reduce  — ~ — ,  — ' -—, — —  ,   ^,  and 

a:4-3        '  a  +  'Zh         '     ^-\-y 

to  intearal  or  mixed  forms. 

a-\-x  ° 

Results,  x^—xy-\-y'^ ^,  a; +  9 -^  or  6  +  2a;  — 

X  -p  y  X  -\-  o 

x^  c 

-,  and  a-\-2b-\ -r^ 

x-{-3'  ^     ^a  +  2b 

4.  Reduce ,  ^  ,  ^    ^  ^ — t?—  ,  and ^ 

a—V    x—y'  a?-{-'2ah  +  l^     '  x-\-y 

to  integral  or  mixed  forms. 

Results,  a-^b,  x^-\-x^y-{-x^y^-i-xy^-\-y^,  a^—x^y-{-xy^—y^, 
and  a^-j-a-{-l. 


161,  Cor. — By  means  of  negative  indices  (exponents)  any 
fraction  can  be  expressed  in  the  integral  form. 

Til  1  1  771 

Thus  —  =  mx-:  and  by  (43)  -  =  ar-\     .-.  —  =  mx-\ 
X  X  ^  ^      ■'  X  X 

5.  Express  -^-^ ,  and =-  in  the  integral  form. 

'  x^y^  mx^  ° 

6.  Show  that  -^4v-o  =  m-%^  +  3/>rVZ»  +  3w-%^ 

m  (a  +  b)~^ 

+  m-^b% 


REDUCTIONS.  Ill 

integral  form. 

Results,  (x-\-y)~^,  a^a^y,  mH^—m^nx^—mn^x^-{-nH^,  and 
7  X  %-hr'y.  

162,  Prob.  3. — To  reduce  numbers  from  the  integral 
OP  mixed  to  the  fractional  form. 

Rule. — Multiply  the  integral  part  by  the  given 
denominator,  and  annexing  the  numerator  of  the 
fractional  part,  if  any,  write  the  sum  over  the  given 
denominator. 

Demonstpation.— In  the  case  of  a  number  in  the  integral  form, 
the  process  consists  of  multiplying  the  given  number  by  the  given 
denominator  and  indicating  the  division  of  the  product  by  the 
same  number,  and  hence  is  equivalent  to  multiplying  and  dividing 
by  the  same  quantity,  which  does  not  change  the  value  of  the  num- 
ber. The  same  is  true  as  far  as  relates  to  the  integral  part  of  a 
mixed  form,  after  which  the  two  fractional  parts  are  to  be  added 
together.  As  they  have  the  same  divisors,  the  dividends  can  be 
added  upon  the  principle  that  the  sum  of  the  quotients  equals  the 
quotient  of  the  sum  (103). 

EXAMPLES. 

Ex.  1.  Reduce  ^a—x^A to  a  fractional  form. 

a—x 

Model  Solution. — Multiplying  2a— x^  by  a— a;,  I  have  20^— ax^ 
—2ax-^x^.    Now  indicating  the  division  of  this  result  by  3a— x',  I 

2a'-aa''—2ax  +  x^  „         ,     dax-Aa""       2a'-ax'-2ax-\-x' 

nave .    .\2a—x^-\ = 

a—x  a—x  a—x 

3^23! 4a' 

H .     But,  as  the  sum  of  these  two  quotients  equals  the 

a—x 

quotient    of    the    sum,    I    have,    after    uniting    similar    terms, 
ar"— Gur'  +  aa:— 2a' 
a—x 

2.  Reduce  a—h-\ j  to  the  form  of  a  fraction. 

Result,  — -v- 


112  FRACTIOITS. 

2x 

3.  Eeduce  1  -\ to  the  form  of  a  fraction. 

y—x 

Result  y^^' 
y—x 

4.  Reduce  a-\-h 7 —  to  the  form  of  a  fraction. 

a—o 

Result, 7,  or  -, 

a—V       b—a 

5.  Reduce  x-\-'Z-\ ^—  to  the  form  of  a  fraction. 

Result. ^' 

x—% 

O/p 5 

6.  Reduce  bx —  to  the  form  of  a  fraction. 

o 

Result,  — 

o 

7.  Reduce  x-{-l-\ to  the  form  of  a  fraction. 

X 

Result,  ^-±^2. 

X 

8.  Reduce  2a— 3^'  + 4c  H ~J~b '  ^' 

J.    T>   1  2abx—2b^x  ,        .      , .  „        x(a~b) 

9.  Reduce  .'c ^—ni —  to  a  fraction.    Res.,  -^ ^. 

a^—b^  a-\-b 

7 


10.  Reduce  x-{-y ^—^ —  Result, 

11.  Reduce  x^  +  %xy  +  f-'^:^^^y^''-t=t. 

Result,  ^l^±f). 
x+y 

d^x  0^7? 

12.  Reduce  -^ -^—{a?x?-\-a}x).  Result,  -, ^• 

13.  Reduce  3«-9- ?^-|?.  Result,      ^ 


a+3  '  «+3 


163,  Prob.  4. — To  reduce  fractions  having  different 
denominators  to  equivalent  fractions  having  a  common 
denominator. 


REDUCTIONS.  113 

Rule. — Multiply  hot) i  terms  of  each  fraction  by  the 
defio7}Unators  of  all  tlie  other  fractions. 

Demonstration. — This  gives  a  common  denominator,  because 
each  denominator  will  then  be  the  product  of  all  the  denominators 
of  the  several  fractions.  The  value  of  any  one  of  the  fractions  is 
not  changed,  because  both  numerator  and  denominator  are  multi- 
plied by  the  same  number  (100). 

EXAMPLES. 

Ex.  1.  Reduce  the  fractions  -,  -^^,  and  --^^  to  eauiv- 

y'  a  +  J  a—h         ^ 

alent  fractions  having  a  common  denominator. 

Model  Solution. — Multiplying  both  terms  of  the  fraction  -  by 
a +6  and  «— J,  or  by  a"— IP,  I  have r^    which  has  the  same  value 

2J 

as  -,  since  the  numerator  and  denominator  have  been  multiplied  by 
the  same  number.    In  like  manner  multiplying  both  terms  of , 

by  y  and  a-5, 1  have  2^^|^iV ,  the  value  of  whic^is 

2—5 
the  same  as = ,  since,  etc.     Finally,  multiplying  both  terms  of 

by  y  and  rt  +  6,  I  have  — ~'~-^- — j;— -,  which  has  the 

same  value  as ^  ?  since,  etc.    These  fractions  have  the  common 

a—b 

denominator  a^y—Vy,  as  in  each  case  the  new  denominator  is  the 

product  of  all  the  old  ones. 

2.  Reduce  ^r->  ^  >  w-^  and  -o  to  forms  havinff  a  0.  D. 
Queries.— By  what  are  both  terms  of--  to  be  multiplied?    By 

what  both  terms  of  —  ?    By  what  both  terms  of  --^? 

2c  Ix 

results,  28c/t2a:y'  2Scn^xy'  28^^y'  '''''^  28^^* 


114  FRACTI01!?^S. 

/>j      iC -I— 1  1 X 

3.  Reduce  ^ ,  — r— ,  and   to  forms  having  a  C.  D. 

n  h 

4.  Reduce  7,  and  — ,  to  forms  having  a  C.  D. 

„      ^^     a^—db        T  al-^W- 
Results,    r, — 75 ,  and      — ^. 

Scholium.— Practically,  this  method  consists  in  multiplying  all 
the  denominators  together  for  a  new  denominator,  and  each  numer- 
ator into  all  the  denominators  except  its  own  for  a  new  numerator. 
But  it  is  much  better  to  repeat  the  rule  as  given  above,  and  let 
that  be  the  form  of  conception,  as  it  keeps  the  principle  constantly 
before  the  mind. 

1G4:»  Cor. — To  reduce  fractions  to  equivalent  ones  having 
the  Lowest  Common  Denominator,  find  the  L.  C.  M.  of  all  the 
denominators  for  the  new  denominator.  Then  multiply  both 
terms  of  each  fraction  by  the  quotient  of  thai  L.  G.  M.  divided 
by  the  denominator  of  that  fraction. 

Demonstration. — The  purpose  in  getting  the  L.  C.  M.  is  to  get 
the  lowest  number  which  can  be  divided  by  each  of  the  denominators. 
That  the  process  does  not  change  the  value  of  the  fractions  is 
evident  from   (100),  the  same  as  under  the  general  rule. 

EXAMPLES. 

a  €?  cti^ 

5.  Reduce  :; ,  -. rr,  and  7- r^  to  equivalent  frac- 

1—a    (l—ay  (1—^) 

tions  having  the  L.  0.  D. 

Model  Solution. — By  inspection  I  observe  that  (1—ay  is  the 
L.  C.  M.  of  the  denominators,  since  it  is  the  lowest  number  which  con- 
tains itself,  and  it  also  contains  each  of  the  other  denominators.   Now, 

to  make  the  denominator  of- ,  (1— a)^  I  must  multiply  it  by 

(1— «)^-f-(l  — a)  ;  ^.  e.,  by  (1  —  ay.    But  to  preserve  the  value  of  the 
fraction,  I  must  multiply  the  numerator  by  the  same  quantity.    Thus 
a    _a(l — <^y _     a — 2a'^  +  a^ 

izia-'ii:^-'-^    .    ..o— ;  etc. 


REDUCTION'S.  116 

6.  Keduce         ^  , to  forms  having  the  L.  C.  D. 

x—y      x-\-y 

Results,  -'-^y^-^y+y\  .^^  t±^±^p:t. 

7.  Reduce  ,  ;.rz~^  "IT  ^^  ^^^^^  having  the  L.  C.  D. 

Results     -^    h(x^-xy^f)  c{x^-xy  +  f) 

8.  Reduce  mn,     ~    ,  — ^^  to  forms  having  the  L.  C.  D. 

m-\-n    m—n 

„      ,,     mhi — mn^    (m — 7iY        ^  (m  +  nY 
Results,  — 7 — ^,  --^ — ^,  and  ^ ^-' 

9.  Reduce ,  - — -^,  - — -3  to  forms  having  the  L.C.D. 

Tlie  L,  a  D.  is  l-\-x—3^—x^. 

10.  Reduce  -, — ^- ,  and  -. 777,  to  forms  having  the  L.  CD. 

a^—¥  {a—hy 

11.  Reduce ,  —z 5, to  forms  having  the  L.  CD. 

m  +  n  m^-\-n^  m-\-n  ° 

12.  Reduce  -,  ^ :; ,  and  -7-^ — -  to  forms  having  the  L.CD. 


/6VT.  Prob.  5. — To  reduce  Complex  Fpactlons  to  the 
form  of  Simple  Fractions. 

Rule, — Multiply  numerator  and  denominator  of 
the  complex  fraction  by  the  product  of  all  the  denom- 
itiators  of  the  partial  fractions  found'  in  them. 

Or,  multiply  by  the  L.  C.  M.  of  the  denominators  of 
the  partial  fractions. 

Demonstration. — This  process  removes  the  partial  denominators, 
since  each  traction  is  multiplied  by  its  own  denominator,  at  least, 
and  this  is  done  by  dropping  the  denominator.  It  does  not  alter 
the  value  of  the  fraction,  since  it  is  multiplying  dividend  and  divisor 
by  the  same  quantity. 


116  FRACTIONS. 


EXAMPLES. 

Ex.  1.  Reduce  -—  to  a  simple  fractional  form. 
Model  Solution. 

Explanation.— In  order  to  free  the  numerator  of  its  denominator, 

2x 
3&',  I  multiply  the  numerator  -^  by   d¥  ;  but  in  order  that  this 

OCT 

may  not  change  the  v^lue   of  the  fraction,   I  also   multiply  the 

denominator  by  the  same.     In  like  manner  to  free  the  denominator 

4v 

-~  of  its  denominator,  I  multiply  numerator  and  denominator  by 

0(1 

5a^   Indicating  these  operations  I  have  j .    To  multiply 

%,x5a''xSl>' 
o« 

--  by  36"  I  drop  its  denominator  and  have  2a;,  which  multiplied 

oO 

by  5a'  gives  for  the  new  numerator  lOa'^x.     So,  also,  I  obtain  the 

new  denominator  by  dopping  5a^  and  multiplying  4y  by  3i^,  getting 

lOa'^x 
thereby  12&'y.    Therefore  the  simple  fraction  is  j^r^-  which  reduced 

to  lowest  terms  is  -^r-. 

2.  Reduce  — —  to  a  simple  form.         Result,  —-- — 

"^  cm 

3.  Reduce  — -|  to  a  simple  fonn.        Result, 


ac+l 

^  '  c 

4.  Reduce — ■  to  a  simple  form.  Result,    ^^ 

x^—  cnx-^cm 

n 


ADDITION.  117 


5.  Reduce to  a  simple  form. 

y 

5 3^ 

6.  Reduce  777 — -~   to  a  simple  form. 

lO  +  ^a;  ^ 

a     X 

7.  Reduce  - — ^  to  a  simple  form. 

m     n 
h—a 


fl  + 


8.  Reduce  uh—a^  ^^  ^  simple  form. 


'HON  IIL 


166,  Prob. — To  add  Fractions. 

Rule. — Reduce  them  to  forms  having  a  comm^on 
denominator,  if  they  have  not  such  forms,  and  then 
add  the  numerators, and  write  the  sum  over  the  com- 
mon denominator. 

Demonstration. — The  reduction  of  the  several  fractions  to  forms 
having  a  common  denominator,  does  not  alter  their  values  (163), 
and  hence  does  not  alter  the  sum.  Then,  when  they  have  a  common 
denominator  (divisor),  the  sum  of  the  several  quotients  is  equal  to  the 
quotient  arising  from  dividing  the  sum  of  the  several  dividends  by 
the  common  divisor,  or  denominator  (103). 


118  FRACTIONS. 


EXAMPLES. 
Ex.  1.  What  is  the  sum  of , :,,  and -• 

Model  Solution. 

Operation.~The  L.  C.  M.  of  1-x,  1—x''  and  1—x^  is  (l—x^) 
x{l+x)  =  l+x—x^—a;\ 

(l+a;)x  (1  +  2x  +  2x'-\-x^)  _  l  +  dx  +  ix''  +  Bx^  +  x* 
(1— a;)x(l  +  2.x  +  2«^+  ic^  "~         1+x—x^—x* 
(\+x'')x(l+x  +  x^)  _  l  +  x  +  2x^  +  x^  +  x* 
{i—x'')x{l+x  +  x^  ~       1+x—x^—x* 

(l+a;")x(l4-g^)  _  1  +  x+x^  +  x* 

(1— «")  X  {1+x)  ~  iTx—x^—x*' 

1+x       l  +  x""       1+x^       l  +  Sx  +  4x''  +  Sx^+x* 


•  l—x   '    1—x'       1—x'  t+x-x^—x* 

l-{-x-\-2x'^  +  x'  +  x'-       1+x  +  x^  +  x*  (A.) 

1+x—x^—x*  1-^x—x^—x* 

~  1+x—x^—x* 

Explanation. — Explain  the  reduction  to  a  common  denominator 
as  under  (163)  unless  that  is  already  sufficiently  familiar. 

Having  reduced  the  fractions  to  the  L.  C.  D,  I  find  (read  A). 

Now  since  the  sum  of  these  quotients  is  equal  to  the  quotient  arising 

from  dividing  the  sum  of  the  several  dividends,  or  numerators,  by 

the  common  divisor,  or  denominator  (103),  I  add  the  numerators 

and  write  the  sum  over  the  common  denominator,   which  gives 

d  +  5x  +  6x''-[-5x^  +  dx*         ^,  ,.  1  +  x  1+x'        ,  1  +  a?' 

'-—-—- ^ r —  foJ*  the  sum  ot , -,  and -„. 

1  +  x—x^—x*  l—x  1—x^  1—x^ 

«     *m    ^     c—a        ,  c-\-a  ^         6x-\-'7c—ff 

2.  Add  ^r-,  — — ,  and— —  Sum,  — 

„     .  ..  a     c     b         ,rt  ^         a^b-rcb-j-il^+-a^ 

3.  Add  Y,  -I,  ~,  and  r-..  Sum,  ^ 

b    ab    a  I^  aW 

4.  Add  3^  ^j  12'  18'  V  ^^^  9'  ^^'^'  ^' 

5.  Add  — r—   and  — ^-  Sum,  — - — 


ADDITION.  119 


6.  Add   :r-— -,   :; ,  lUld  :— —  •  Sum, 

l+«    1— «           l+fl  I— a 

7.  Add .  and   -- — -„•  aSmw,  - — ^. 

8.  Add  - — -  and  , Sum, 


\-\-x           1-^x                                      '  1— a^ 
9.  Add  -5 — ^  and  7 — -775-      /S'wm,  -^ ^y iTTTS* 

10.  Add  ::^-,-f-  ^»d  — — • 
a^—y^   x-\-y  ^—y 

Sum,  ?^f-te. 

:/67.  Cor. — Expressions  in  the  mixed  form  may  either  be 
reduced  to  the  improper  form  and  then  added,  or  the  integral 
parts  may  he  added  into  one  sum,  and  the  fractional  into 
another,  and  these  results  added. 

Ex.  1.  Add  7a:+^  and  Sa;4-^^- 
o  ox 

17B8T  FOBM  OF  OFBBATION. 

«— 2       22a;— 2 

7x  -I- 

8a;  + 


3  3 

3a;+4       40x'  +  3a;+4 


5a;  5x 

(22x-2)x5a;  _  llOx'-lOx 
3      X  5.r  ""  \5x 

(40x'4-3x+4)x3  _  1^20^H9«  +  1_2 
5a?  X  3  ~  15x 

llOx'-lOx       120^  jf  ?^  13  __  230a;''-x4-12 
15x  15x  ~  15a; 

SECOND  FOBM  OF  OPERATION. 

7.r  +  8x  =  15x 
(a;— 2)  X  5a5  _  5x'— lOx 

3      X  5x  15x 

(3x  +  4)x3_  9ic  +  12 

5x      X  3  ~     15aj 


120  FRACTIONS. 


...  7.  +  —  +  a.  +  ^-  =  15X+  _--  +  ,—   =  15,,. 

"^  15x 

Explanation. — Since  the  sum  of  several  numbers  is  the  same  in  what 
ever  order  their  parts  are  added,  I  may  add  the  integral  parts  first. 
Adding  7.r  and  Sx  I  have  1 5x.   Reducing  the  fractions  to  forms  havin;^- 

a  common  denominator, ' becomes  — -- — -,  and  — - —  becomes 

;j  15;r  5x 

— — —  .     Adding  these  I  have   .'  ~ which   added    to   15x, 

the  sum  of  the   integral    parts,   gives  for   the   entire    sum    15.r 

5a;'— a;+12 
■*"        15^        • 

2.  Add  2x,  3.^■-f-— ,  and  x-\--^-  iSum,  ^^-^tt' 

«     Ajj         Sa^  ^     .     2ax        .,  ,  ,  ,  2abx—3cx^ 

3.  Add  a 5-  to  Jh ^^<w,  a  +  o-\ 7 

be  DC 

4.  Add  7a;  H — ^r-  and  9a; r 

3  ^x  5:i'2— 16a;4-9 


iS'w?w,  16a; -f 


15.T 


_     ,  , .  ^       3a;    2a;     ^  ,   .,       a;        ^^  17a; 

5.  Add  6a;— J,  -^ — 8a;,  and  3a;—--      xS'?m,  x — —^ 

6.  Add  3ma;— ^^,  and  ^^— 2ma;  +  4. 

4aJ 

7.  Add  6a;%i— 3a;— ;§^  and  5a;-2a;%*  + 


jr?$'       jor  qr 

.     .  , ,      a          3a          ,         2aa;  <.  ^ 

9.  Add  — —  ,  — —  ,and  — x ,•  Sum,  — — -- 

a— a;    a  +  a;  a^— .t^  a-\-x 

Suggestion.— The  L.  C.  D.  is  (a—h)  Q)—c)  (c—a). 


;V—      -Sy^— 


JOS.  Prob. — To  subtract  Fractions. 

Rule. — Reduce  the  fractions  to  forms  having  a 
c^jmivon  denominator,  subtract  the  numerator  of  the 
subtrahend  from  the  numerator  of  the  minuend, 
and  place  the  remainder  over  the  common  denomi- 
nator. 

Demonstration.— Since  the  values  of  the  fractioua  are  not  altered 
by  reducing  them  to  forms  having  a  common  denominator,  their 
difference  is  not  altered.  Now  subtracting  the  numerator  of  the 
subtrahend  from  the  numerator  of  the  minuend,  and  dividing  the 
difference  by  the  commcm  denominator,  gives  the  diflFerence  between 
the  fractions,  since  the  difference  of  the  quotienttt  of  two  quantities 
divided  hy  a  common  divisor,  is  the  xame  as  the  quotient  of  the  difference 
divided  by  the  same  divisor  (104). 


EXAMPLES. 

Ex.  1.  From  ^^±^  subtract  ^^^> 
x—y  x+y 

Model  Solution. 
Operation,     (x—y)  (x-\-y)  =  x^—y"^ 

{x-\-y)x(x+y)  _  x'  +  2xy+y* 
(x-y)x{x  +  y)  x^-y^ 

(x—y)  X  (x—y)  _  a;'— 2.ry  +  y» 
(x-\-y)x{.r—y)  x^—y^ 

(x'  +  ^.ry^  2/-^._(.r»— 2  ry  -f-  f)  =  4xy 

G 


122 


FRACTION'S. 


x  +  y      x—y  _    ^xy 
x—y      x  +  y      x'^—y' 

Explanation. — The  L.  C.  M.  of  x—y  and  x+y  is  their  product,  since 
tbey  have  no  common  factor.    Hence  x^—y"  is  the  L.  C.  D.    To  reduce 

X  +  11 

— ^to  a  form  having  this  denominator,!  multiply  both  its  terms  bv 
x—y 

x^  +  2a?y  + 11^ 

x+y^  which  gives y~^  t     -     ^^  like  manner  multiplying  both 

X     y 

x-^y  ^  XI         x'^  —  2xy  +  y'^       _, 

tenns  of by  x—y,  I  have  — v-   o      .     I  have  now  to  subtract 

x+y  x'—y^ 

/J.2 ^xv  +  v'^  x'^  +  2^  +  v^ 

^   \       from ^      „  ^  .     Since  the  difference  of  the  quotients 

x^—y^  x^—y^  ^ 

of  two  numbers  divided  by  the  same  number,  is  the  same  as  the 

quotient  arising  from  dividing  the  difference  between  those  numbers 

by  the  common  divisor,  I  take  the  difference  of  the  numerators  (the 

quantities  to  be  divided)  which  is  4a^,  and  dividing  it  liy  x'^  —  y^  I 


have 


ixy 


x  +  y  X- 

for  the  remainder  of  — -  less  — 


2.  From  ^-^   take 

l-\-x 

1 

a-\-x 

x-\-3 


3.  From 

4.  From 


1—x 

1 

a — x 

X 


x—3 
5.   From  dx  take 


take 
take 


X 

Sa-{-12x 


Remainder, 
Remainder, 
Remainder, 
Remainder, 


4:X 


1—x^ 
2x 

q2 — ^2 
9__ 

x^—3x 
3x—Sa 


6.  From  dy  take  ^-^• 
1      ,  ,  1 


7.  From 

8.  From 


take 


a^—2ab-hb^ 


Remainder,  — ^^ — 

o 

a—b—1 


a — i 

_1_  ^^^^      ^—x^—3x 
0^  +  4    ^^^  x^^\fdx-\-%^ 


Rem., 
Rem., 


{a-hf 


.    ^         3x  +  2  ,  ,      lax—AOa      „        .    ,       4(3-.r) 
9.  From take Remainder,  -^ <- 


SIBTRACTIOX.  133^ 

34.22;       2— 3a; 


10.  Combine    the    following    fractions 


2—x        2-\-x 


H 5 — 7-'  Result,  -— --■ 

^    aH— 4  x-\-2 

1  1 

11.  Combine 


a  {a  —  b)  {a  —  c)        b  {b  —  c)  (b  —  a) 

+  —, w IT-  Result,  -J-' 

c(c  —  a){c  —  b)  abc 

,^    ^,      ,.       Sa—^b       2a—b—c       15a— 4c       a—ib 
12.  Combine  —^ +  -^ ^^    • 

Sla-4:b 


Result, 


84 


13.  Combine  — -^  + 


a-\-b   '    rt2__52       a^^b^ 


Result,  —. — :,-' 

14.  Combme ^-  —  .--rr^ 7-5 — r*       Result,  0. 

-2a;       14- 2a;       ^7?—\  ' 

15.  Combine  ^t-t — — -r  —  ^ttt-, r^  —  ^-7;^ — --^  • 

2(a;  +  l)       10  (a;— 1)       5(2a;4-,3) 

Result,  - 

16.  Combine  -^ 5  + 


;.r2-l)  (2a; +  3) 


2f?.9.  Cor. — Mixed  numbers  may  be  subtracted  by  annexing 
the  sid)lraJiend  with  its  sign^  changed,  to  the  minuend,  and  then 
combining  the  terms.  The  reason,  for  the  change  of  signs  is  the 
same  as  in  ichole  numbers  (77). 

EXAMPLES. 

17.  From  3a;  +  ?i^  take  x  -  ^^. 
m  m 

2x 

Remainder,  2x  -\ • 

m 


iU 


FRACTIONS. 


18.  From  x jr—  subtract  7a; 


Remamder,  —  -  -—  6a;  —  a. 

0 


19.  From  'da  take 


dx  +  l)la 


Remainder^ 


3a— Sx 


20.  From  2a;  4 ^r—  take  3a; -^ — 

Remainder, 


16a; +  23 
42 


ultinlicatinn 


ECT10N  ¥= 


170.  Prob.  1. — To  multiply  a  Fraction  by  an  Integer. 

Rule.  —  Multiply  the  numerator  or  divide  the 
denominator. 

Demonstration. — Since  numerator  is  dividend  and  denominator 
divisor,  and  the  value  of  the  fraction  is  the  quotient,  this  rule  ia  a 
lirect  Nionsequence  of  (101,  102). 

EXAMPLES. 

Ex.  1.  Multiply  -X by  m-{-n. 

Model  Solutlpit 
Operation.     ^— x(m+«)  = -^^  . 

Explanation. — Since  the  value  of  a  fraction  is  the  quotient  of  the 
numerator  divided  by  the  denoc.iinator,  and  multiplying  the  divi- 


MULTIPLICATION.  1^5 

dend    multiplies    the    quotient,   I    multiply    the    numerator    of 

^~     bym  +  w,  and  have  — ;; ,  which  is  therefore  m+»timeg 

m—n 
3xy   • 

2.  Multiply  1^  by  3«. 

Model  Solution. 

Operation.    3^,jiX3«=^r- 

Explanation.— Since  27W«  is  to  be  divided  by  3<z'ft'  if  I  divide  the 
divisor  by  3a,  thus  making  it  da  times  as  small,  it  will  go  into  the 

27/12?  . 

dividend  3«  times  as  many  times  as  before.     Hence  ~j^  is  3a  times 
--,^  .     The  operation  is  performed  by  cancellation. 

3.  Multiply  1^  by  x-\-y. 


^'  M^^^ipiy  Ti^y)  ^y  ^-^ 


5.  Multiply  ^-  by  a^+i.  p^^^.^      ^ 


Suggestion,    a^—a  =  a(a*—\)  —  a{a^—l)  (a'  +  l). 

6.  Multiply  -^^^^  by  (a+y)». 

Suggestions. — Multiply  by  one  of  the  factors  of  (a +y)'  by  rejeot- 

ing  it  from  the  denominator,  giving  — r ,  and  this  product  by 

(«  +  V)' 

the  other  factor,  giving  ^^ . 

am — my 

3  * 

7.  Multiply by  a^—a?.  Prod.,  Sa  +  Sa:;. 

8.  Multiply  -g,  by  2a+2y.  Prod.,  .^. 

9.  Multiply  ^-^^-  by  3J.  Proef.,  3a-2y. 
Bimply  remove  the  denominator.     Why  ? 


136  FRACTIONS. 


10.  Multiply  by  x—y. 

x—y 

11.  Multiply  by  ^ax  {x—y).  Prod.,  6a^a^, 

12.  Multiply  -^  by  x^-1. 

X  —  1.  ' 

13.  Multiply  -"t^  by  x^-2xy+y\ 

X — y 


171,  Prob.  2.— To  multiply  by  a  Fraction. 

Rule. — Multiply  by  the  numerator  and  divide  by 
the  denominator. 

Demonstration. — Let  it  be  required  to  multiply  m,  which  is  either 

an  integer  or  a  fraction,  by  =~  • 

1st.  Suppose  a  and  b  are  both  integers.     Multiplying  m  by  a 
gives  a  product  h  times  too  large,  since  we  were  to  multiply  by  only 

a  5th  part  of  a;  hence  we  divide  the  product,  am,  by  &,  and  have  -j-. 

2nd.  When  either  a  or  5,  or  both,  are  fractions.     Let  c  be  the 

factor  by  which  numerator  and  denominator  of  ^  must  be  multiplied 

to  make  =-  a  simple  fraction  (165).  Then  will  tt  be  a  simple  frac- 
tion, i.  e.,  ac  and  he  are  each  integral;  and  the  multiplication  is 

effected  as  in  Case  1st,  giving  -—  •     This  reduced  by  dividing  both 

oc 

terms  by  c,  i.  e.,  by  cancelling  c,  gives  -r-  •  Hence  we  see  that  in 
any  case,  to  multiply  by  a  fraction,  we  have  only  to  multiply  the 
multiplicand  by  the  numertitor  of  the  multiplier,  and  divide  this 
product  by  the  denominator.  It  is  also  to  be  observed  that  this 
reasoning  applies  equally  well  whether  the  multiplicand  is  integral 
or  fractional. 

EXAMPLES. 


Ex.  1.  Multiply  (^-^^  by  ^. 


MULTIPLICATION.  12, 


Model  Solution. 


Operafon.    -,_,  x  (^-y)  =  ^.^^^^ 
,      .        {a—Vf  ,^  a-h 


{a—hf      x—y  _       a—'b 


x^—y^      a— I      x^-vxy-vy"^ 

Explanation.— In   order  to   multiply  -^ ^   by  ^|,   I  first 

X  —  y  (I — 0 

multiply  by  x—y.    This  is  effected  by  dividing  the  denominator, 

j.3_y9    bv  x—iL  (IO*i),   and   <nves   — •     But,   since   this 

multiplier  was  to  be  divided  by  «— 5,  the  product  now  obtained 

must  be  divided  by  the  same.     Dividing  — r  by  a—h  by 

^  x'  +  xy  +  y^ 

dividing  the  numerator  (101),  I  have  for  the  complete  product 

a— J) 
x^+xy-ity"" 

Practimlly,  the  operation  is  performed  by  cancelling  x^y  from 
«'— y',  leaving  x^  +  xy  +  y"^ ;  and  a—h  from  (a— 6)^,  leaving  a—h. 

2.  Multiply  5  by  |.  Prorf.,  ^. 

3.  Multiply  ^j^  by  -y  Prod.,  ^^-^. 

A    TiM  li.-  .         Sex  ,      2c  „     ,        2c2.r 

4.  Multiply  -g^^  by  3^.  Prorf.,  -^. 

6.  Multiply  -3^^  by  -^^^.  Pro</.,  ^j^- 

Scholium. — When  there  are  no  common  factors  in  the  numerators 
and  denominators  of  the  fractions  to  be  multiplied  together,  the 
process  consists  simply  in  multiplying  numerators  together  for  a 
new  numerator  and  denominators  for  a  new  denominator. 

6.  Multiply  ^x^-9y^  by 


2x^3y 


128  FRACTIONS. 

Suggestion. — In  this  case  divide  first,  obtaining  3a? -f-3y.     Multi- 
ply this  by  2m. 

Prod.,  imx  +  Qmy. 

7.  Multiply  -  by  — —  Prod.,  -^- 

^  -^   a    ''  a-\-c  a^-{-((c 

8.  Multiply  -^  by  -^-.  Pro^.,  p^^- 

9.  Mulfply  ^^^  by  ^-j-  Prorf.,  ^^;^jv 

Suggestion. — Jn  the  last  example  both  operations  are  performed 
upon  the  denominator  of  the  multiplicand. 

10.  Multiply  -^-  by  ^^-.  Pr.<;.,  ^^-g- 

11.  Multiply  ^  by  3^^-p^.  Prod.,  j^- 

12.  Multiply  '^hy^.  Prod.,  "^^. 

13.  Multiply  M;  by  j^;.  P.od,  ^. 

14.  Multiply  -  by  -•  Prod.,  — „■ 

15.  Multiply  ^  by  ^.  Pro(^.,  ? 

16.  Multiply  —. — 7--  by  — —-  Prod.,  -^i- 

17.  Multiply  —  by -^ ,    and  write    the    result 

6a^y  ^    *"    7«~3^  8 

without  negative  exponents.  p     ,    Qcx'^y 


MULTIPLICATION.  129 


18.  Multiply  together  -^,  _(£tJ'?,  and  _^^^. 


Prod., 


a*  {x-y) 

19.  Multiply  -^^:^i  by  ^:p^^2- 

17^.  Cor. — To  multiply  mixed  numbers  first  reduce  (hem 
to  improper  fractions. 

20.  Multiply  1-g  by  ^-2. 

Prod     y^y-2^-6l^   ^^.^17yz-M_a 

1—^2    1— V^  '»• 

21.  Multiply  together  — — - — ,  -^— ^,  and  14- 


l+y  '  r^  +  ic'  1— a; 

X 

22.  Multiply  «--  by  -+-•  Prod.,  ^^. 

(I         X     a  (IX 

23.  Multiply  l-^II^  by  2  +  -^.       Proe?.,  -|^. 

24.  Multiply  x^-x-\-l  by  i^^-^l. 

3J^        X 


Prod.,  a^+lH-^" 


Operation. 

£c*  —  a;  +  1 

a;       or 


a; 
«»  -«  +    1 


+    1  +  -, ,  Prodttct, 


130 


FRACTIOI^^S. 


17 S*  Prob.  1. — To  divide  a  Fraction  by  an  Integer. 

Rule.  —  Divide  the  numerator  or  midtiply  the 
denominator. 

Demonstration. — Since  numerator  is  dividend,  and  denominator 
divisor,  and  the  value  of  the  fraction  the  quotient,  this  rule  is  a 
direct  consequence  of  (101,  102). 


EXAMPLES. 

Ex.  1.  Divide  — -^  by  a—x. 

Solution.— Since  the  value   of  this  fraction  is  the  quotient  of 
3(a*— a;^)  divided  by  2mx—l,  if  I  divide  the  dividend,  3(a'— a?'),  by 


3(a"-«5 
2mx—l 


{a-x)  = 


3a  +  Sx 

2mx—l 


a—x,  I  divide  the  fraction.     Hence 

2.  Divide  —^^^  by  a-\-b. 

Solution. — Since  the  value  of  this  fraction  is  the  quotient  of 

dm— 4x1/  divided  by  a—h,  if  I  multiply  the  divisor,  a—h,  by  a  +  &,  I 

v  •  1    xu    i?     X-  TT  dm—4xy       .      ,.       3m—4xy 

divide  the  fraction.     Hence j-^  h-  (a  +  6)  =  — . — ^r-  • 

a—h  a^-h^ 

3.  Divide  -r-^  by  3a;. 

4o?h      ^ 

4.  Divide  by  5(«  +  a;),        Quot^  -^ -• 

x^ 1 

5.  Pivide  ^  ,  ^  by  x—l, 


^+3 


DlVlSIOIf.  131 


6.  Divide —  by  x—y, 

x^  ——  u^ 

7.  Divide  ^  by  a:  +  y. 

8.  Divide  -^  by  ^H4. 

9.  Divide  ^"^  by  ^- vl 

x+y     ^  ^ 

^    ■'  (^+y)^ 

Suggestion. — Perform  this  example  by  dividing  successively  by 
the  factors  of  x^—y\  viz.,  x^-y  and  x—y.  To  divide  by  «— y, 
divide  the  numerator;  to  divide  by  «  +  y,  multiply  the  denominator. 

10.  Divide  -^^  by  5mnW.  G^o^.,  s?^* 

11.  Divide T^ —  by  a^—d^. 


12.  Divide  ^l^f    ^^-  by  42  (a-^>)2. 


174.  Prob.  2. — To  divide  by  a  Fraction. 

Rule. — Divide  by  the  niuiierator  and  multiply  the 
quotient  by  the  denominator. 

Or,  what  is  the  same  thing,  invert  the  terms  of  the 
divisor  and  proceed  as  in  multiplication. 

Demonstration. — The  correctness  of  the  first  process  appears  from 
the  fact  that  division  is  thereverseof  multiplication,  which  requires 
that  we  multiply  by  the  numerator  and  divide  by  the  denominator. 

The  process  of  inverting  the  divisor  and  then  multiplying  by  it 
is  seen  to  be  the  same  as  the  other,  since  this  also  multiplies  the 
dividend  by  the  denominator  of  the  divisor  and  divides  b^  the 
numerator. 


132  FRACTIOlSrS. 

Again,  this  process  may  be  demonstrated  thus  :  Inverting  the 
divisor  shows  how  many  times  it  is  contained  in  1.  Then  if  the 
given  divisor  is  contained  so  many  times  in  1,  it  will  be  contained 
in  n,  n  times  as  many  times;  or  in  any  dividend  as  many  times  the 
number  of  times  it  is  contained  in  1,  as  is  expressed  by  that  divi- 
dend, whether  it  be  integral,  fractional  or  mixed. 

Scholium  I. — Since  to  multiply  one  fraction  by  another  we  may 
multiply  the  numerators  together  for  the  numerator  and  the  denom- 
inators for  the  denominator,  and  since  division  is  the  reverse,  we 
may  perform  division  by  dividing  the  numerator  of  the  dividend  by 
the  numerator  of  the  divisor,  and  the  denominator  of  the  dividend 
by  the  denominator  of  the  divisor. 

This  method  will  coincide  with  the  others  when  they  are  worked 
by  performing  the  operations  by  division  as  far  as  practicable,  and 
this  is  worked  by  performing  the  multiplications  equivalent  to  the 
divisions  when  the  latter  are  not  practicable. 

EXAMPLES. 

Ex.  1.  Divide  ^^,  by  — f-. 

Model  Solutions. 

OPBBATION  BT  THB  FIRST  METHOD. 

x^  +  y' 

2y 


X'  +  ': 


X  («+y)  =  iJ 


Explanation.— I  first  divide  ^-^  by  y,  by  dividing  the  numera- 

x^+y 

tor  (101).     But  the  given  divisor  is  y  divided  by  x  +  y;  and  as 
dividing  the  divisor  multiplies  the  quotient  (102),  I  must  multiply 

this  quotient, -J-—,  by   x-\-y.     Performing  this  by  dividing  the 
X  •\-y 


denominator,  I  have  for  the  true  quotient  -^ -„ 

'  ^  x^—xy  +  y^ 

OPERATION  BT  THB  SEOOKB  METHOD. 

y    __    ^''   ^  x^y  _        2y 


x^+y^  '  x  +  y      x^+y^     '  y        x^—xy+y"" 


biVisiox.  133 

Explanation. — By  inverting  the  divisor  and  indicating  the  multi- 
plication of  tiie  dividend  by  it,  I  indicate  that  the  dividend  is  to 
be  multiplied  by  xA-y  and  divided  by  y,  which  are  the  opei-ations 
required,  lu  this  instance  I  perform  the  multiplication  by  ^-f-y,  by 
dividing  the  denominator,  and  the  division  by  y,  by  dividing  the 
numerator. 

The  operation  by  the  third  method  is  of  the  same  form  as  the  last. 

BXFLAKATION  BT  THS  THIRD  KBTHOD. 

I  am  to  find  how  many  times  -^  is  contained  in  ~^-.      Now 

x^y  aj'  +  y" 

-^—  is  contained  in  1,  — -  times:  since  y  is  contained  in  1,  - 

«+y  y  y 

1  X  '\-'U 

times,  it  is  contained  mx-\-y.x-{-y  times  -,  or  — -  times.    Hence  if 

y       y 

V     .  .     .  .    .    x+v   .         .    .  .     ,  .       2^^^         2v* 

-^—  IS  contained  m  1,  — -  times,  it  is  contained  m    ,  ^    „  -r^—^ 
«4-y  y  x'+y''  x'+y' 

y 


times  —^ 


2.  Divide  ,    .    ,„  by  -^ — .•  Q'^^*-^      /     ,     \' 

«    ^.  .J     4  (a^—ab)  ^      Qab  /,    ^    2  (a— by 

4.  Divide  «-+3a^+3..^  +  ^  ,    J^^ 

Quot,  — i— . 
x—y 

3  ^^"'^     12a: 


6.  Dmde  — ^  by  -^-  C^o^., 


fl    T^-  -^        ^^     u        ^  ^     /     2a;(a;  +  l) 

6.  Divide  ^-— ^  by  _•  q^ot,  -^^. 

8.  Divide  -^£7.  I^v  '^.  C^o^.,  1. 

a*— 2«d4-62    "     a^b  a 


134  tRACTION^S. 

9.  Divide  3^---^^  by  j--^.  Qmt.,  ^-J^. 

,-     ^.  .,  1x     .  2x  ^     ,     x—\ 

10.  Divide  x-\ ~   by  x Quot., -- 

x—d     *'  x—^  ^  x—5 

11.  Divide  xf^ •.  by  x • 

x^     -^          X 

Suggestion. — This  quotient  should  be  written  by  inspection  in 
the  same  manner  as  {x*—y*)-7-{x—y),  and  is  x^  +  x  +  -  +  —■     Or 


it  may  be  performed  as  follows 


1 

X 

X 


X* 

— -  x^  +  X  -\-  -  +  - 

-1  XX* 

X* 

x-"  -  1 


L  _  L 

X^  X* 

L  _  L 

X^  X* 


12.  Divide  j-~  +  j^  by  f^-j~^-         Quot.,  I. 
Suggestion. 

1  X      _l+x''  _1 X      __  1+x* 

1+x'^  \—x  ~  1—x^  1—x     1+x  ~  1—x^' 


13.  Divide  — ^  +  — -^  by  —^— -^• 

a;— ^^       x-\-y    ''  x—y      x-\-y 

«-■■  ^- 


DirisioK.  135 


16.  Divide :— ^  by  -7; ?< 


Quoty  3x^  —  xy  —  2y2. 

Scholium  2. — It  is  sometimes  conveniedt  to  write  the  divisor 
under  the  dividend  in  the  form  of  a  Complex  fraction,  and  then 
reduce  the  result  to  a  simple  fraction  by  (165). 

16.  Divide  1-f-  by  1—^. 


L      aj  (T  +  d  a 

Operation. 


[-yx"'  " 


a=»_l       a-i 


Operation. 


(ib*  +  a¥—ab* 


U  +  ^-llxaJ" 

_         &=>  +  !         _*+l 
-a5='(6»-6+l)  ~  a5«  ' 

18.  Divide  Trhj—if  by  xy-^-^x'^y. 


3      .     r^,-2/^1xxy      ,    3 

y—y     _  La;-         J _  y'— x'g 


Suggestions. 3 r-  =  :i ^ =  —. 5  • 

19.  Free     ,3  '^  _^  from  negative  exponents. 

Result,  — ^— Ts-i* 
axy^-\-l^y^ 

d-s J-2 

20.  Free    _^     ,  _3  from  negative  exponents. 


a^b^  +  a^ 


136  FRACTIOKS. 

21.  Free  — Tjf-o  ^^^"^  negative  exponents. 


mn~*e 


I ^-2 yZ 

22.  Free r-    ^    '    -^  from  negative  exponents. 

^^•'  ~^^-a^-ha^xy^  ' 

23.  What  is  the  reciprocal  of  --^^J 

Solution. — The  reciprocal  of  a  quantity  being  the  quotient  of  1 
divided  by  that  quantity,  the  reciprocal  of  -^— ^  is  1-^  -.r"^2>  ^' 


17o,  Scholium  3. — Thus  it  appears  that  the  reciprocal  of  a 
fraction  is  the  fraction  inverted. 

The  reciprocal  of  a  quantity  with  a  negative  exponent  is  the 
same  quantity  with  a  positive  exponent. 

-2./-S  x^ifi 


24.  What  is  the  reciprocal  of — ^^-r-  Ans.,  -f-- 

25.  What  is  the  reciprocal  of  7 ~^,? 

(a—b)-' 

[To  be  taken  in  review.] 

26.  Multiply    ^~-  by  ^-  •  Prod.,    ^--. 

«» — lA        a^—h*  a»—o» 


SYNOPSIS. 


137 


SYNOPSIS    FOR    REVIEW. 


z 

O 


2i 

<  I 

fiC    I 


it 


o 

"Si 


FRACTION. 


FORMS. 


(  Terms, 
NUMBRATOB.        Value  of  fraction. 

J  Cot.  1. — Multiplying  or  dividing 
Denomxmatob.  I         numerator  or  denominator. 

I  Cor.  2.— Removing  denominator. 


Fractionai..     i  Proper.        J 
Mixed.  I  Improper     \ 


Simple. 

Compound. 

Complex. 


Lowest  terms.— L.  C.  D.— Redaction. 

Of  numerator,  of  denominator,  effraction. 

Essential  sign  effraction. 

PROB.  1.— To  lowe&t  termt<.    Rule.    Bern.    Sch.    By  H.  C.  D. 

PROB.  2.— Prom  Improper  to  integral  or  mixed  forms.    Ruuk 
Dem.     Cor.    Negative  exponents. 

PROB.  3.— Prom  integral  or  mixed  to  fractional  forms.  Rule.  Dem 

PROB.  4.— To  common  denominators.       Rule.       Dem.        Cor. 
L.  C.  D.     Dem. 

PROB.  5.— Complex  to  simple.    Rule.    Dem. 

ADDITION.    Pkob.    Ritle.    Dem.    Cor.    Mixed  Numbers. 

SUBTRACTION.    Pbob.    Rule.    Dem.     Cor.    Mixed  Numbers. 

(Pbob.  1.— Fraction  by  integer.    Rule.    Dem.    Sc/i. 
MULTIPLI-   3 


CATION. 


Pbob.  2.— Any  number  by  fractions.    Rule.    Dem. 
{  Sr-h.    '\>r.    Mixed  Numbers. 


(Pbob.  1.— Fractions  by  integer.    Rule.    Dem. 
DIVISION.     -nProb.  2.— An V  number  by  fraction.    Rttlb.    Dem». 
(  1,2,8.    5Wi*.  1,2,3, 


Test  Questions. — Upon  what  five  principles  in  Division  are  moat 
of  the  operations  in  fractions  based?  Why  does  the  process  of 
reducing  to  forms  having  a  common  denominator  not  change  the 
value  of  a  fraction  ?  Give  the  rules  for  Multiplication  and  Division 
of  Fractions,  and  the  reasons  for  them. 


c//; 


lY. 


A^^ 


Section  i 


INVOLUTION. 

The  operations  in  Radicals  are  all  based  upon 
the  most  elementary  principles  of  Factoring.* 

If  the  student  learns  how  to  use  this  key,  he  can  unlock  all  the 
mysteries  of  the  subject. 

GENERAL    DEFINITIONS. 
170.  A  Power  'n  a  produd  iirisiug  from  multiplying 

a  number  by  itself. 

The  Degree  of  the  power  is  indicated  by  the  number  of 
factors  taken. 

Thus  2,  4,  8,  16,  and  33  are,  respectively,  the  1st,  3nd,  3d,  4th, 
and  5th  powers  of  3. 

Scholium. — It  will  be  seen  that  a  power  is  a  species  of  comi3osite 
number  in  which  the  component  factors  are  equal. 

177,  A  Root  is  one  of  the  equal  factors  into  which  a 
number  is  conceived  to  be  resolved. 

The  Degree  of  the  root  is  indicated  by  the  number  of 
required  factors. 

*  The  eubjects  treated  in  this  chapter  are  among  the  most  difficult,  if  not  actually 
the  most  difficult  for  the  pupil  in  the  whole  science.  In  the  examination  of  hundreds 
of  students  from  all  parts  of  the  country,  the  author  has  found  that  the  rule  is  that 
they  are  deficient  in  knowledge  of  Badicals.  An  attempt  is  here  made  to  assist  the 
teacher  in  remedying  this  defect,  by  constantly  holding  tha  attention  to  this  one 
central  principle. 


INVOLUTION.  lS(i 

Thus,  2  is  the  1st  root  of  2,  tlie  2nd  root  of  4,  the  3rcl  root  of  8, 
the  4th  root  of  16,  the  5th  root  of  32,  etc. 

178,  Scholium  I. — P<w^  and  i2(x>/ are  correlative  terms.  Thus 
32  is  the  5th  power  of  2,  and  2  is  the  5th  root  of  32. 

170.  Scholium  2. — The  Second  Power  ia  &\so  c&Wed  the  Square  ; 
the  Third  Power ^  the  Cube ;  and,  sometimes,  the  Fourth  Power ^  the 
Biquadrate.  In  like  manner  the  2nd  root  is  called  the  square  root ; 
the  third  root,  the  cube  root ;  the  fourth  root,  the  biquadrate  root. 
These  are  Geometrical  terms  which  have  been  transferred  to  other 
branches  of  mathematics.  The  second  power  is  called  the  square, 
because,  if  a  number  represents  the  side  of  a  square,  its  second 
power  represents  its  area,  or  the  square  itself.  Conversely,  if  a 
number  represents  the  area  of  a  square,  the  square  root  represents 
the  side.  Also  the  third  power  represents  the  volume  of  a  cube,  the 
edge  of  which  is  the  first  power  or  cube  root.  Biquadrate  means 
twice  squared,  and  hence  the  fourth  power. 

180.  An  Exponent  or  Index^  is  a  number  written  a 
little  to  the  right  and  above  another  number,  and  indicates 

1st.  If  a  Positive  Integer,  a  Poiver  of  the  number  ; 

2nd.  If  a  Positive  Fraction,  the  numerator  indicates  a 
'Poiver,  and  the  denominator  a  Root  of  the  number; 

3d.  If  a  Negative  Integer  or  Fraction,  it  indicates  the 
Reciprocal  of  what  it  would  signify  if  positive. 

Illustration.  4'  is  the  3rd  power  of  4,  or  64.  a"'  is  the  with  power 
of  a,  if  m  is  an  integer.     4"'  (read  "  4,  exponent  -3  ")  is  j^,  or  — . 

.        1  2 

a-"»  IS  --.    8*  is  the  cube  root  of  the  square,  or  the  square  of  the 

cube  root  of  8,  or  4.     a»  is  the  mth  power  of  the  nXh  root  of  a,  or 
the  nth  root  of  the  mth  power,  if  m  and  n  are  both  integers.  a~^<  is  — . 

Scholium. — It  is  obviously  incorrect  to  read  4t,  "  the  ^  power  of 
4."  There  is  no  such  thing  as  a  2-fifths  power,  as  will  be  seen  by 
considering  the  definition  of  a  power.    Read  4l,  "  4  exponent  | ; " 


140  POWERS  AKD   ROOTS. 

m  m 

also  a»,  "  a  exponent  ^  "  ;  a  »',  "a  exponent  —  f."  These  are  abbre- 
viated forms  for,  "  a  with  an  exponent  — f ,"  etc.  In  this  way  any 
exponent,  however  complicated,  is  read  without  difficulty. 

181.  Cor. — A  FACTOR  can  be  transferred  from  numerator 
to  denominator  of  a  fraction,  or  vice  versa,  bij  changing  the 
sign  of  its  exponent,  without  altering  the  value  of  the  fraction. 

Thus  ^'"^"  =  ?!^-  for  ^'"^~"  —  __?!  _^_cr     y »  _  «V 

182.  A  Radical  Number  is  an  indicated  root  of  a 
number.  If  the  root  can  be  extracted  exactly,  the  quantity 
is  called  Rational ;  if  the  root  can  not  be  extracted  exactly, 
the  expression  it  called  Irrational,  or  8urd. 

Thus  the  radical  ^25a"  is  rational,  but  ^^\Za  is  surd. 

183.  A  Root  is  indicated  either  by  the  denominator  of  a 
fractional  exponent,  or  by  the  Radical  Sign,  y'.  This 
sign  used  alone  signifies  square  root.  Any  other  root  is 
indicated  by  writing  its  index  in  the  opening  of  the  v  part 
of  the  sign. 

Thus  '^am,  \^am,  are  the  3rd  and  5th  roots  of  «m,  and  the 
same  as  (aw)i,  («w)i. 

184.  An  Imaginary  Quantity  is  an  indicated  even 
root  of  a  negative  quantity,  and  is  so  called  because  no 
number  taken  an  even  number  of  times  as  a  factor,  produces 
a  negati\  e  quantity. 

Thus  \/—4:  IS  imaginary,  because  no  number  multiplied  by 
itself  once  produces  —4.     Neither  +2  nor  —2  produces  —4  when 

squared.     For  a  like  reason  y^— Sa'*,  \/—5x,  or  y^ —14:0xy^  are 
imaginaries. 

ISS,  All  quantities  not  imaginary  are  called  Real. 


INVOLUTION.  141 

1S6.  Similar  Radicals  are  like  roots  of  like  quantities. 
Tims  4  V^j  '^^  V^">  ^^^  {a^—x^)  \/5a  are  similar  radicals. 
IS't.  To  Rationalize  an  expression  is  to  free  it  from 
radicals. 

ISS,  To  afifect  a  number  with  an  Exponent  is  to 
perform  upon  it  the  operations  indicated  by  that  exponent. 

Thus  to  affect  8  with  the  exponent  |  is  to  extract  the  cube  root 
of  the  square  of  8,  or  to  square  its  cube  root,  and  gives  4. 

1S9,  Involution  is  the  process  of  raising  numbers  to 
required  powers. 

li)0.  Evolution  is  the  process  of  extracting  roots  of 
numbers. 

191,  Calculus  of  Radicals  treats  of  the  processes  of 
reducing,  adding,  subtracting,  or  performing  any  of  the 
common  arithmetical  operations  upon  radical  quantities. 


INVOLUTION. 

192.  Prob.  1. — To  raise  a  number  to  any  required 
power. 

Rule. — Multiply  the  tiitniber  by  itself  as  inany  times, 
less  one,  as  there  are  units  in  the  degree  of  the  power. 

Demonstration. — Since  the  number  of  factors  taken  to  produce  a 
power,  is  equal  to  the  degree  of  the  power  (1T6),  it  follows  that  to 
obtain  the  2nd  power  we  take  two  factors,  or  multiply  the  number 
by  itself  once;  to  obtain  the  3rd  power  we  take  three  factors,  or 
multiply  the  number  by  MoeXHwke ;  and  in  like  manner  to  obtain 
the  nth  power  we  take  n  factors,  or  multiply  the  number  by  itself 
n— 1  times. 

EXAMPLES. 
1.  What  is  the  3rd  power  of  'Za^  ? 


142  POWERS   AND   ROOTS. 

Model  Solution. — Since  the   3rd  power  of  2a^  is  the  product 
arising  from  taking  it  3  times  as  a  factor,  I  have  2a'  x  3a'  x  3a' 

=  8a^ 

2.  What  is  the  4th  power  of  —lOah 

(_10«i)  X  (-lOa^)  X  (-  lOai)  x  (-  10«4) 
=3  lOOOOa^,  Answer. 

3  to  6.  What  is  the  square  of  —bmhi  ?    Of  6aH  ?     Of 

--a-n?    Of-—? 
5  5^>2   • 

a^       9        9 
Answers,  257nhi^,  S6j- ,  — ^--  ,  ^r^a^b-^. 

7  to  10.  What  is  the  cube  of  —2al^?     Of  12^"^?     Of 

_^^..     Of-^% 
3  6n 

Answers,  —Sa^¥,  172S7rr^-  =  ^^ ,  -^.a^  ^* 


^yit  27      '      216n3 

11.  What  is  the  squareof  2a— 3a;?  or,  {2a— Sx)  x  (2«— 3^:)? 

12.  What  is  the  3rd  power  of  2d^^Zx  ? 

Ans.,  8rt«-f-36a^a;  +  54a2a:2  +  27a:3. 

13  to  17.  Expand  the  following:    {\-^2x-^%^f,   {l^x 

j^x'^—^Jf,  (a-^b—cf,  (l+2a;  +  :^'")^  and  {\~^x^'6x'^—7?f. 

Results,  l-{-4tx  +  l^x^  +  12x^-\-^a^,  l—2x-\-dx^—4.a^-\-Zx^ 

—2x^  +  x^,  l  +  Qx^lbx^  +  2^x^-^lbx^-itQ7^  +  x^,  and   l  —  iSx 
4-15a;2-20a;3^15^_6^_^^6. 

1Q      Q!U  (27a4_18a2Z,2_J4)2       (9,,2_52)3(^^_^2) 

18.  Show  that  ^ ^-^^ 1  +^ 64^^^—  =^' 

19.  What  is  the  square  of  9ar  +  -  ?    ^ns. ,  81a^2  ^  is  _(.  _  . 

20.  Expand  as  above  (4—^^)1        Result,  16-8^7^  +  ^7, 

21.  Expand  (a^-\-c^Y. 

Eestilt,  a^-\-4:ah^-i-6ac-}-4:ah^-^c\ 


INVOLUTION.  143 

193.  Cor. — Since  any  number  of  positive  factors  gims  a 
positive  product,  all  powers  cf  positive  monomials  are  positive. 
Again,  since  an  even  number  of  negative  factors  gives  a  posi- 
tive product,  and  an  odd  numbei-  gives  a  negative  product, 
il  follows  that  even  powers  of  negative  numbers  are  positive,  and 
odd  powers  negative. 


104,  Prob.  2.  —  To  affect  a  monomial  with  any 
exponent. 

Rule. — Per  form  upon  the  coefficient  the  operations 
indicated  hy  the  exponent,  and  multiply  the  exponents 
of  the  letters  hy  the  given  exponent. 

Demonstration. — 1st.    Wheri  the  exponent  in  a  positive  integer. — Let 

it  be  required  to  affect  4a'"6"aj-*  with  the  exponent  p ;  or  in  other 
words  raise  it  to  the  pth  power. 

It  is  obvious  that,  if  ^nH'x-'  is  taken  p  times,  each  of  its  factors 
will  be  taken  p  times. 

Now  4  taken  p  times  as  a  factor  is  represented  thus,  4** 

Again,  a"'  taken  p  times  as  a  factor  becomes a'^P 

because  a'"  signifies  a  taken  m  times  as  a  factor,  and  if 
this  group  of  m  factors  is  taken  p  times,  a  will  have 
been  taken  mp  times,  giving  a'^P. 

n  1 

In  like  manner,  &~  signifies  &'  taken  as  a  factor  n 
times,  and  if  this  grouj^  of  n  factors  is  taken  p  times, 

V  will  have  taken  np  times,  making pr 

Lastly,  X—  signifies  —  (180,  3d) ;  and  this  taken  as 

IP 

a  factor  p  times  give  — .     But  as  f  =  1,  this  fraction 

is  by  (1§1) x-P* 

Collecting  the  factors  we  have ...............  ^Pa"^Pb  *"  x-p^ 

Q.  E,  D, 


144  POWEKS   AND   ROOTS. 

2nd.   When  the  exfponeni  u  a  positive  //'action.     Let  it  be  required 

to  affect  4:a"'b'x-%  with  the  txponeiit  '       This  means  that  4a'"5~ar-» 

is  to  be  resolved  into  q  equal  factors  and  j9  of  them  taken.  Or  that 
we  are  to  find  the  qth  root  of  the  quantity  and  then  raise  it  to  the 
joth  power  (180,  2d).     If  we  then  separate  each  of  the  factors  of 

n 

4a»»&Ta!-*  into  q  equal  factors,  and  then  take  p  of  each  of  these,  wi 
shall  have  done  what  is  signified  by  the  exponent  -• 

By  definition,  one  of  the  q  equal  factors  of  4  is 4:5 

Also  by  definition,  one  of  the  q  equal  factors  of  a«»,  or 

m 
the  ^h  root  of  a"*  is  represented  ( 1 83)  by «« 

To  separate  ft"  into  q  equal  factors,  we  notice  that  &~ 
is  n  of  the  r  equal  factors  of  h.  Now,  if  we  resolve  each 
of  these  r  factors  into  q  equal  factors,  h  is  resolved  into 
rq  equal  factors;  doing  the  same  with  each  of  the  n  fac- 

.  tors  represented  by  5~,  and  taking  one  from  each  set,  we 

2L 

have ft**« 

which  is  therefore  one  of  the  q  equal  factors  of  J>~. 

To  resolve  x-*  =  —  into  q  equal  factors,  we  consider 

that  a  fraction  is  resolved  by  resolving  its  numerator  and 
denominator  separately.    But  one  of  the  q  equal  factors 

of  1  is  1 ;  and  o?ie  of  the  q  equal  factors  of  x"  h  X9  as  seen 
in  the  resolution  of  a"\     Hence  otie  of  the  q  equal  factors 

11  -i 

of  or*,  or  — ,  is  —  = x  « 

a;'       •- 

X9 

Collecting  these  factors  we  find  that  one  of  the  q  equal 

«  1     pi    n         s 

factors  of  4taH~  ar-*  is 4:«ay  h''*ix   7 

And  finally  p  of  these  being  obtained  according  to  Case  1st,  gives 
4:9a  'i'brqx  ?,  as  thc  expression  for  A:aH~x-^  affected  with  the  expo- 
nent - ;  which  result  agrees  with  the  enunciation  of  the  rule. 


INVOLUTION.  145 

3rd.   When  the  exponent  is  negative  and  either  integral  or  fractional. 

u 

Let  it  be  required  to  affect  Aa'"h~'jr*  with  the  exponent  —t.  This, 
by  definition  of  negative  exijoiients,  signifies  that  we  are  to  take  the 
rcoii^rocal  of  what  the  ex|vcssi(>n.  would  be  if^  were  positive.     But 

4a"'&'ar-'  aftected  with  the  exponent  t  (positive)  is  4:'a'"'b^x-*% 
whether  t  is  integral  or  fractional,  as  shown  in  the  preceding  cases. 

The  reciprocal  of  this  is .     But  since  these  factors  can 

be  transferred  to  the  numerator  by  changing  the  signs  of  their  expo- 
nents, we  have  4-'a-""J  ~aj",  as  the  result  of  aflFecting  4a'"5'ar-*  with 
the  exponent  —t,  which  result  agrees  with  tl»e  enunciation  of  the 
rule. 


[Note. — The  above  demonstration  contains  the  fundamental 
principles  of  the  whole  subject  of  the  Theory  of  Exponents,  and  it 
is  of  the  highest  importance  that  it  be  made  perfectly  familiar. 
The  application  of  the  rule  is  so  simple  in  practice  as  to  afford  no 
diflSculty  ;  but  the  reasoning  should  be  given  in  full  in  a  suflicient 
number  of  the  following  examples  to  jvx  in  the  mind  these  principles 
of  the  theory.  After  this  is  done,  the  pupil  needs  only  to  perform 
the  operations.  The  danger  is  that  the  liow  being  so  simple,  the 
why  will  be  disregarded.] 


EXAMPLES. 

Ex.  1.  Affect  2a2jfc-*  with  the  exponent  5:  that  is,  raise 
it  to  the  5th  power. 

Model  Solution. 
Operation.  {M'lic-y  =  32a'°&nrc-". 

Explanation.     {2a^h^c-*y  is  2a''l>h-*  x  2a^l)^<r-*  x  2a«5l<r-* to 

6  factors.  This  gives  5  factors  of  2,  or  32 ;  5  factors  of  a'  or 
10  factors  of  a,  a'" ;  5  factors  of  5if,  which  I  obtain  by  considering 
that  6t  is  2  factors  of  fti,  and  hence  5  factors  of  rf  Ig  5  times  2  fac- 
tors of  ji,  or  b^ ;  and  since  c-*  =  —  ,  5  factors  of  it  are  —  x  —  -  -  to 


146  POWERS   AND   ROOTS. 


5  factors  =  -^  =  (r-^\  as  fractions  are  multiplied  by  multiply- 
ing   numerators   and   denominators.       Hence   I   have   (3a^6lc~*)* 

2.  Affect  3a^  with  the  exponent  m ;  that  is,  raise  it  to  the 
w?th  power,  m  being  an  integer. 

Explanation.    (Sal)'"  =  3at  x  3a^  x  3o« to  m  factors.     This 

gives  m  factors  of  3  and  m  factors  of  a^.  But  m  factors  of  3  are 
represented  by  3**.  To  obtain  m  factors  of  a^^  I  consider  that  at  is 
2  factors  of  a^ ;  hence  m  factors  of  it  are  2m  factors  of  ak.  That  is, 
a  resolved  into  3  equal  factors  and  2m  of  them  taken.    Therefore 

2  ^ 

3.  Affect  16a%~*  with  the  exponent  f ;  that  is,  represent 

2  of  the  3  equal  factors  of  16a%~^. 

Explanation. — I  will  first  resolve  IBa^^iC-"  into  3  equal  factors  (or 
indicate  it  when  I  cannot  perform  it).     To  do  this  I  take  one  of  the 

3  equal  factors  of  16,  which  I  represent^  as  I  cannot  resolve  it,  and 
write  (16)^.  Again,  one  of  the  3  equal  factors  of  «^^  is  «^,  as  a*^  is 
15  factors  of  a,  and  consequently  when  resolved  into  3  equal  factors 
one  of  the  3  contains  5  factors  of  a.  Thus  axaxa to  15  fac- 
tors when  put  into  3  equal  groups  becomes  aaaaa  x  aaaaa  x  aaaaa^ 

one  of  which  is  a\    To  resolve  x-^  I  consider  itas—  =  -x-x- 

X^         X       X       X 

X  -  X  -  X  - .   Combining  this  into  3  equal  groups  it  becomes  -^  x  — 

XXX  XX 

X  —  ;  hence  one  of  the  3  equal  factors  of  -  is  -, ,  or  ar-".   Therefore 
«''  ^  x^     x^' 

16a^^xr^  being  resolved  into  3  equal  factors,  one  of  thwn  is 
(lQ)^a''xr-^.  But  I  am  to  take  2  of  these,  as  the  exponent  f  indicates. 
This  gives  me  (repeat  the  reasoning  of  Ex.  1)  (16)la^''c-*. 

4.  Affect  da'^b-^  with  the  exponent  J  ;  that  is,  take  m  of 
the  n  equal  factors  of  it. 


INVOLUTION.  147 


Explanation.     3"  represents  one  of  the  n  equal  factors  of  3.     One 
of  the  n  equal  factors  of  a"  is  a,  as  a"  means  n  factors  of  a.    h-^ 

=  —  =  -X  ^x  ^x torn  factors.   Each  of  these  being  separated 

into  n  equal  factors,  I  have  -[^"1^1: ^^^  factors  =  - .     And 

in         J^         ^n  0 

taking  one  of  the  factors  -j- ,  from  each  of  the  m  factors  =- ,  I  have  m 

factors  of  —  or  —  =  6~^.     Therefore  one  of  the  n  equal  factors  of 
6"      hi 

^n^-m  ig  3»a5— ;.    But  I  am  to  take  m  such  factors  which  gives  (re- 

m  m 

peat  the  process  of  Ex.  2)  3  "a"*  6   «. 

5.  What  is  the  square  of  -ahn  ?   Of  —  ^r-^-  *  or  —-abx-^  ? 

The  cube  of r ,  or akwrkz  ? 

3ml  3 

2 

6.  Affect  8a%~*  with  the  exponent  5*    Result,  4a8ar*. 

o 

4 

7.  Affect  32a»2^i«  with  the  exponent  — -• 

o 


Result, 


8.  Affect  13a;~»2/~^  with  the  exponent  —3. 

Result,  -^^^f- 

9.  Perform  the  following  operations  and  explain 
each  as  a  process  of  factoring:  (^%a^b~^)~^,  (lOOa"^^^)^, 
(lla2a;-iy-2)-i,  and  {a^b^x-»)-l 

Results,  — ^,  10  -  ,  -— ^  ,  and  --- . 


148  POWERS   AND   ROOTS.- 

195.  Prob.  3.— To  expand  a  binomial  affected  with  any 
exponent. 

Rule. — This  rule  is  best  stated  in  a  formula.  Thus, 
let  Si,  h,  and  m  be  any  numbers  ivhatever,  positive  or 
negative,  integral  or  fractiortal,  then  will  (a-j-b)™ 
represent  any  binomial,  affected  with  any  expo- 
nent, and 

•        1  •  -4 

^  m(m-l)  (m-2)^^_,^, 

m{m-l)  (m-2)  m-^) 
+  1.2.3.4  ""      ^ 

m{:m-l)  (m-2)  (m-3)  (^-4)       . 
+  1.2.3.4.5  "^      ^4-  etc. 


Demonstration.  —  This  formula  is  the  celebrated  Binomial 
Theorem  discovered  by  Sir  Isaac  Newton.  There  are  several 
elementary  demonstrations,  one  of  which  will  be  found  in  Appen- 
dix I.  If  the  pupil  learns  the  formula,  and  learns  to  apply  it  with 
facility,  it  is  all  that  is  thought  best  for  him  to  attempt  at  this 
stage  of  his  progress. 


EXAMPLES. 
Ex.  1,  Expand  {x-\-yy  by  the  Binomial  Formula. 

Model  Solution. — To  apply  the  B.  F.,   x  =  a  of  the  formula, 

y  =  6,  and  5  =  wz.     .-.I  have    {x  +  yT  =  x^  -\-  5x^-^y  -\ — - 

,  ,  ,       5  (5  -  1)  (5  -  2)    ,  ,  ,      5  (5  -  1)  (5  -  2)  (5  -  3)    ,,  , 

■\ — ^    ~  ~        A~    K~     ic^~Y,  at  which  term  the  develop- 


INVOLUTION.  140 

ment  becomes  complete  since  the  next  coefficient  would  have  a 
factor  5—5  =  0  which  would  destroy  the  term.  Performing  the 
tiperations  indicated  I  have  (x  +  yY  =  3f -\- ox*i/ +  lOx^y^ -\- lOx^y^ 
+  5xy*  +  y'\  (In  practice,  this  result  should  be  written  out  with- 
out writing  the  preceding,  by  simply  applying  the  formula 
mentally.) 

2.  Expand  (x—yY  by  the  B.  R 

Suggestions,     x  =  a,    —y  =  ft,    and    6  =  w.      .'.    (x—yY  =  ar" 

^Qa^i-y)  +  ^^x<i-yy+^-^x^(-yr  +  ^-^:lx^-yy 

6-5-4-3-2     ,      ,,      6-5'~4-3-2    1^,        ~         ,      ^, 
+ jg x(-yy  + ar»  (  -  y)^  =  a:«  -  Qx'y 

+  15a;y-20arV+  15a;Y-6ajy^  +  y". 

3.  Expand  (2m2— 3^^)*  by  the  B.  F. 

Suggestions. — Make  a  =  Sw",  h  =  —  3/i*,  and  w  =  4,  and 
(2w»-3wi)*  =  167^^"— 96w^•ni  +  216m*w-2167n='wt  +  8l7l^ 

4.  Expand  (x-\-y)-*  by  the  B.  F. 

Suggestions.  a  =  x,b  =  y,  and  m  =  —4.  This  series  does  not 
terminate,  since  no  factor  ever  becomes  0 ;  but  the  development 
can  be  carried  to  any  desired  extent.  (x  +  y)-*  =  x-*  —  4ar-'y 
+  lOar-V  —  20ary  +  35a^V  — ,  etc. 


5.  Expand  (m  —  n)^  by  the  B.  F. 

Suggestions. — Making  the  proper  substitutions  in  the  formula, 
we  have  (m  —  ny  =  m^  +  |w^~^  (—  n)  +  ^^\     ^  w*~*  (—  ny 


etc.  A  series  which  never  terminates,  since  no  coefficient  reduces 
to  0.  Performing  the  operations  indicated  we  find  that  (wi— n)f 
=  mi—^m~^n—^7nrin'*—^m~in*—^m~^n*'-,  etc. 

*  [S^ia  read  "  factorial  3,"  and  means  the  product  of  the  natural  numhers  from  1 
to  8  inclusive.    |5_i9  the  same  as  1  x  2  x  .3  x  4  «  5,  etc. 


150  POWERS   AND   ROOTS. 

6.  Expand  (1  +  a;)^  by  the  B.  F. 

r>      7.    /I        x„       ,              n(n—l)  ^     7i(n—l)  (n—2)  „ 
Result,  (i-^xy'=:  i-^nx  +  -^ — ^-x^-{ — ^^ f^ -V 

Scholium. — TLis  expansion  is  in  itself  a  very  useful  formula,  and 
should  be  memorized. 


7.  Expand  (3-?/2)i  by  the  B.  F. 

Result,     (3-2/^)i  =  3i-f^2^2_p^_3^2/« 

-y^—,  etc. 


128 

8.  Expand  (l-\-x^Yhj  the  B.  F. 

Result,  1  +  5a:2_^  10^:4  _|_i0a;6^5;^^^  ^10. 

9.  Expand  (l—a^)-i  by  the  B.  F. 

13  5  35 

Result,  l4--a2+-a4+_a6_^__a8^^  etc. 


10.  Expand  Va^—a^e^  by  the  B.  F. 
Result,    Va^—a^e^  =  aVl^^  =  a{l—e^)^  =  «(1— h  ^ 
1    .     1.3    .      i-3:i_^_  ^^  . 

Qp" — ,  etc.;. 


2.4        2.4-6        2.4-6.8 


11.  Expand  ^^^  by  the  B.  F. 


(c-\-xy     ^      ^ 

1  ,,      2a:     3ic2     4a:s 
+  etc.  =  -(!_--  +  —__  +,  etc.). 


IKVOLUTIOK.  151 

12.  Expand  — ^ — r  by  the  B.  F. 

H ,  etc. 

^625  ' 

I 

13.  Expand  (x  +  y  +  c)*  by  the  B.  F. 

Suggestions.— Put  (x -{■  y)  =  z.  .-.  (x  +  y  +  c)*  =  («  +  c)*  =  «* 
+  42V  +  62V  +  4«c'  +  c*  =  (restoring  the  value  of  z)  (x  +  y)* 
+4(x4-y)''c  +  6(x4-y)V  +  4(a5  +  y)c»  +  c\  But  (.r  +  y)*  =  a;*+4a;'y 
+  6xy  +  4.ry^  +  y*,  (j-  +  y)«  =  r3  +  3x'y  +  3xy'  +  y^  and  (x  +  y)-^  =  :c^ 
^-2xy^-y^  Whence  by  substitution  we  have  (x  +  y  +  c)*  =  x*  +  4x''y 
+  6xy  +  4xy'  +  y*  +  4cx«  +  12cxV  +  12cxy'  +  4cy»  +  6c'V  +  12c''ajy 
+  6cy  +  4c'x  +  4c='y  +  c*. 

14.  Expand  (2a—b-{-(^Y  by  the  B.  F. 

Result,  Sa^  —  12a^  -\- 6ai^  ^  I^ -\-  12aV  —  12«^c2  ^  3j3c3 

196,  Cor.  1. — Uie  expansion  of  a  binomial  terminates 
only  when  the  exponent  is  a  positive  integer,  since  only  when 
m  is  a  positive  integer  will  a/actor  of  the  form  m(m— 1)  (m— 2) 
(m— 3)  etc.  become  0. 

197.  Cor.  2. — When  m  is  a  positive  integer,  that  is  when 
a  binomial  is  raised  to  any  power,  there  is  one  more  term  in  the 
development  than  units  in  the  exponent.  Since  the  first  coeffi- 
cient is  1  ;  the  2nd,  m  ;   the  3rd,  -^ ;   the  4th, 

m{m-l){m-2)_  ^^^  ^^^^  m(,m-l)  (m-^)  (m-3) ,  ^^^  ^ 

we  notice  that  the  last  factor  is  m —  (the  number  of  the 
terni  —2) ;  and  the  number  of  the  term,  therefore,  which 
has  m—m  as  a  factor  is  the  (7?i  +  2)th  term.  But  this  is  0. 
Hence  the  (m  +  l)th  term  is  the  last. 


152  POWERS  AND  ROOTS. 

198,  Cor.  3. —  When  mis  a  positive  integer  the  coefficients 
equally  distant  from  the  extremes  are  equal ;  since  (a  -f  by* 
—   (J  _l_  a)^ ;  the  former  of  which  gives  a^  +  ma"^^ 

^i7i|m— ^^^_2^2_|.^  etc.     Whence  it  appears  that  the  first 
half  of  the  terms  and  the  last  half  are  exactly  symmetrical. 

199,  Cor.  4. — The  stem  of  the  exponents  in  each  term  is 
the  same  as  the  exponent  of  the  power. 

Scholium. — The  last  two  corollaries  apply  to  the  form  (x-^-y)"^, 
and  not  to  such  forms  as  (2a^  —  35^)'",  after  the  latter  is  fully 
expanded. 

200,  Cor.  5. — A  contenient  rule  for  writing  out  the 
POWERS  of  binomials  may  he  thus  stated  : 

1.  The  FIRST  term  contains  only  the  first  letter  of  the  binomial, 
and  the  last  term  only  the  second,  while  all  the  other  terms  con- 
tain  both  the  letters. 

2.  The  exponent  of  the  first  letter  of  the  binomial  in  the  first 
term  of  the  development  is  the  same  as  the  exponent  of  the 
required  potver  and  diminishes  by  unity  to  the  right,  while  the 
exponent  of  the  second  letter  begins  at  unity  in  the  second  term 
of  the  expansion  and  increases  by  unity  to  the  right,  becoming, 
in  the  last  term,  the  same  as  the  exponent  of  the  power. 

3.  The  coefficient  of  the  first  term  of  the  expansion  is  unity  ; 
of  the  second,  the  eocponent  of  the  required  j^ower  ;  and  that  of 
any  other  term  may  be  found  by  multiplying  the  coefficient  of  the 
preceding  term  by  the  exponent  of  the  first  letter  in  that 
term,  and  dividing  the  product  by  the  exponent  of  the  second 
letter  +1. 

4.  There  is  one  more  term  in  the  development  than  there  are 
units  in  the  exponent  of  the  power. 


INVOLUTION.  163 

This  rule  is  a  deduction  from  the  formula  («  +  &)'"  =  a'"  +  ina"'-'b 
m  (m-1)  w(^-l)(^-2)      _        wi(m-l)(7n-2)(m-3) 

a'"-*6* etc. 

The  1st  point  appears  from  Cob.  8. 

The  law  of  the  exponents  is  directly  observable  from  the  formula. 

The  coefficients  of  the  first  and  second  terms  are  seen  in  the 
formula  to  be  as  stated.     The  coefficient  of  the  third  term  may  be 

written  m  x  — - — ,  which  is  the  coefficient  of  the  second^  or  preced- 

ing  term  (m),  multiplied  by  the  exponent  {m—\)  of  the  first  letter 
in  that  term,  and  divided  by  2  which  is  the  exponent  (1)  of  the 
second  letter  +  1.     In  like  manner  noticing  any  other  term,  as  the 

5th,    its    coefficient    may     be    written     — ^^ ^^- ^  x  — -— . 

[3  4 

But  — ^^ —- '  is  the  coefficient  of  the  4th  term,  m—S  is  the 

exponent  of  the  first  letter  in  the  4th  term,  and  4,  the  divisor,  is  the 
exponent  of  the  second  letter  (3)  -1-  1. 

The  4th  point  appears  from  Cor.  2. 

16  to  20.  Write  out  by  the  above  rule  the  expansions  of 
the  following :  (m  +  ny,  (x  +  y)^,  (a  -f-  ^ )^  (x  -f-  m)%  (iri^  +  ^3)3^ 

Suggestions  upon  the  last. — Regard  a^  and  rm  as  simple  num- 
bers represented  by  letters  without  exponents.  Thus  (a^  +  w^)* 
=  (al)«  +  4  {a^y  {m^)  +  6  (J)'  (m^)'  +  4  {a^)  (/7ii)'  +  (mi)*.  Now 
performing  the  operations  indicated,  we  have  (a»  4-  mir)*  =  af 
+  4a»mi  +  6a^m + iaJni  +  m'. 

201.  Cor.  6. — If  the  sign  between  the  terms  of  the  bino- 
mial is  minus  and  the  exponent  is  a  pof-itive  integer,  as  (a — b)''*, 
the  odd  terms  of  the  expansion  are  +  and  the  even  ones  — . 
This  arises  from  the  fact  that  the  odd  terms  involve  even  powers 
of  the  second  or  negative  term  of  the  binomial,  and  the  even 
terms  involve  the  odd  powei^s  of  the  same. 


154  POWERS  AN^D   ROOTS. 

Thus  the  second  term  involves  (—5)  which  makes  its  sign  -- ; 
the  4th  term  has  (—5)',  the  6th  term  (-&)',  etc.  But  the  first  term 
does  not  involve  (— &),  and  the  3rd  has  {—bf  or  h'\  the  5th  has 
{-by  or  b\  etc. 

21  to  24.  Write  out  the  expansions  of  {m—ny,  (x^yy, 
Result  of  the  last,  {a^—h^y  —  a2_4alji  ^  ^db^—^c^l  +  it 


EVO  LUTI  0  N. 

202.  Prob.  1. — To  extract  the  ^th  (any  root)  of  a  per 
feet  power  of  that  degree. 

Rule. — Resolve  the  numher  into  its  prime  factors, 
and  separate  these  into  m  equal  groups;  one  of  these 
groups  is  the  root  sought. 

Demonstration. — Since  the  wvth  root  (i.  e.,  any  root)  of  a  num- 
ber is  one  of  the  m  equal  factors  of  that  number,  if  a  number  is 
resolved  into  m  equal  factors,  as  the  rule  directs,  one  of  them  is  the 
mtla.  root. 

EXAMPLES 

Ex.  1.  Extract  the  cube  root  of  74088. 

Model  Solution. — Resolving  74088  into  its  prime  factors  I  find 
them  to  be  2-3-3-3-3-3  7  •  7  •  7.  These  arranged  in  3  equal 
groups  give  3-3-7x2-3-7x2-3-7.  Hence  3 •  3  7  =  42  is  the 
cube  root  of  74088,  since  it  is  one  of  the  3  equal  factors. 


2  to  5.  Extract  in  this  manner  the  following :  V 492804, 
'V^592704,  ^248832,  ^456533.      Roots,  702,  84,  aad  12. 

6.   Extract  the  square  root  of  Sla^x~~^yiz~^. 

Model  Solution. — The  two  equal  factors  of  81  are  9-9;  of  a* 
a^a^;    of  x-^,  ar^-x-^-^    of  ys,  y^  yi  ;    of  z~^^  z~ts  •  s~tV,     Hence 


EVOLUTION.  166 

8ia*x-^y^z~^  =  Qd'x-^y'z'^o  x  Qa^ar-^y^z~TG^    and    consequently, 
9«'a^'y»«~iff  is  its  square  root. 


V: 


7  to  11.  Extract  \/2la^f^,  ^Uar^x^,  V^9xy^^  ^/U^ahn^ 
f~-    Roots,   ±ba%^,  ±%a-H^,   ±7ic%i,   ±\2ahn% 

12  to  15.  Extract   \/l25m^x^,  "^ll^Sx^y^,  \^—32a^^y-^y 

Boots,  5?n^x^,  VZx^^,  —2a^y-\  and  ±2n-^y\ 
Query.— Why  the  ambiguous  sign  to  the  last  ? 

203.  Scholium. — The  sign  of  an  even  root  of  a  positive  num- 
ber is  ambiguous  (that  is,  +  or  — )  since  an  even  number  of  factors 
gives  the  same  product  whether  they  are  positive  or  negative  (87, 
§8).  The  sign  of  an  odd  root  is  the  same  as  that  of  the  number 
itself,  since  an  odd  number  of  positive  factors  gives  a  positive 
product  and  an  odd  number  of  negative  factors  gives  a  negative 
product  (88,  89). 

204,  Cor.  1. — The  roots  of  monomials  can  be  extracted  by 
extracting  the  required  root  of  the  coefficient  and  dividing  the 
exponent  of  each  letter  by  the  index  of  the  root,  since  to  extract 
the  square  root  is  to  affect  a  number  with  the  exponent  ^,  the 
cube  root  ^,  the  nth  root  i,  etc.   (194.) 

EXAMPLES. 
16  to  21.  In  this  manner  write  V^25«^^,  v^— 343aSr«, 

^^^^-  ^^^"^'  /S'  -^  \/iS^- 


243%- 

3a^x^ 
Roofs,   ±6ah,    -'7x^y-%    ±3mhi'-^x^,   3(^m-^y^,    ^7,, 

2a^^ 
and — 

3itw-2 


156  POWERS    AND    ROOTS. 

205.  Cor.  2. — The  root  of  the  product  of  several  numbers 
is  the  same  as  the  product  of  the  roots.  Thus,  \/ahcx 
=  v^a  •  \^b  •  v^c  .  V^:r,  since  to  extract  the  mth  root  of  abcx 
we  have  but  to  divide  the  exponent  of  each  letter  by  m, 

which  gives  ah^c^x^  or  v^a  •  v" J  •  v^  •  v^. 

206,  Cor.  3. — The  root  of  the  quotient  of  two  numbers  is  the 
same  as  the  quotient  of  the  roots.    Thus,  A  /  —  is  the  same  as 

— —  ,  since  to  extract  the  rth  root  of  —  we  have  but  to  ex- 

"'  n 


n 
tract  the  rth  root  of  numerator  and  denominator,  which 

operation  is  performed  by  dividing  their  exponents  by  r. 


TT  r/m       m^        \/m 

Hence  \  /  -  =  — r  =  — 

V  ^         71'        a7 


EXAMPLES. 

22.  Show  that  ^8"^^  =  v^8  x  v^27. 

Model  Solution. — We  may  show  this  in  two  ways.  1st. — Experi- 
mmtally.  Thus  ^8x27  =  ^216  =  6.  Again  v^8  x  v^27  =  3x3 
=-.6.  Hence  ^^8^^  (or  the  cube  root  of  the  product)  =  ^8  x  v^27 
(or  the  product  of  the  cube  roots).  2nd. — Analytically.  \^^  x  27 
signifies  that  the  product  of  8  and  27  is  to  be  resolved  into  3  equal 
factors,  which  is  accomplished  by  resolving  8  into  its  prime  factors, 
and  27  into  its  prime  factors,  and  then  separating  these  factors  into 
three  equal  groups  (202).  This  will  give  the  same  result  as  resolv- 
ing the  product  of  8  and  27,  or  216,  into  3  equal  factors,  since  the 
prime  factors  of  216  are  the  same  as  those  of  8  and  27. 

23.  Show  that  '^'ar-'^  =  ^^-  x  ^^. 

1  _m    2.         _2   J. 

Suggestions. — The  5  equal  factors  of  «-""&»  are  a  ^&6~  a~'^b^% 

m     1         _m     l_  -Jl    L 

oT^t^,  a  "^b^^  and  a  ^h^",  since  by  the  rules  of  multiplication  these 


EVOLUTION.  167 


1 


multiplied  together  make  «-*"&".    But  a  *&"•  is  the  product  of  one 
of  tlie  5  equal  factors  of  a-""  by  one  of  the  5   equal   factors  of 


5^  or  ^f^  by  j/ft*. 


24.  Extract  the  square  root  of  a^^-\-^d^M-\-aW^(?, 

Solution. — The  factors  of  this  are  readily  seen  to  be  a',  c',  and 
a''  +  2a^  +  &',  which  separated  into  two  equal  groups  givea<^a  +  &) 
and  ac(a  +&).     Hence  ac{a  +  &)  or  a^c  +  (the  is  the  required  root. 

25.  Extract  the  square  root  of  m'^—'lm^x-\-m'^:i^. 

Root,  tf?  {^—x)  or  m^—mhi» 

Scholium. — The  extraction  of  roots  by  resolving  numbers  into 
their  factors  according  to  this  rule,  is  limited  in  its  application  for 
several  reasons.  In  the  case  of  decimal  numbers  we  can  always  find 
the  prime  factors  by  trial,  and  hence  if  the  number  is  an  exact 
power,  can  get  its  root.  But  in  case  the  number  is  not  an  exact 
power  of  the  degree  required,  we  have  no  method  of  approximating 
to  its  exact  root  by  this  rule,  as  we  have  by  the  common  method 
already  learned  in  arithmetic.  In  case  of  literal  numbers  the  difficulty 
of  detecting  the  polynomial  factors  of  a  ])olynomial  is  usually 
insuperable.  Hence  we  seek  general  rules  which  will  not  be  sub- 
ject to  these  objections. 


207.  Prob.  2. — To  extract  the  square  root  of  a  poly- 
nomial. 

Rule. — /.  Ai^ange  the  polynomial  with  reference 
to  one  of  its  letters,  as  for  division. 

II.  Extract  the  square  root  of  the  first  left  hand 
term.  Hiis  root  is  the  first  term^  of  the  required  root. 
Subtract  the  square  of  this  teim  of  the  root  from  the 
polynomial. 

III.  Double  the  root  already  found  for  a  Trial 
Divisor.  By  this  trial  divisor  divide  the  first  term  of 
the  remainder  of  the  polytwmial,  and  write  the 
quotient  as  the  second  term  of  the  root. 


158  POWERS    AKD    ROOTS. 

IV.  Complete  the  divisor  by  adding  to  the  trial 
divisor  the  last  term  in  the  root.  Multiply  the  Ti^ue 
Divisor  thus  formed  by  the  last  term  in  the  root,  and 
subtract  the  product  from  the  last  remainder,  bring- 
ing down  such  terms  as  may  be  necessary. 

V.  Repeat  the  process  of  dividing,  completing  the 
divisor,  multiplying  and  subtracting,  in  the  same 
may  till  the  polynomial  is  exhausted,  or  until  thepe  is 
no  term  of  it  remaining  which  can  be  exactly  divided 
by  the  first  term  of  the  trial  divisor. 

Demonstration. — 1st.  The  polynomial  is  arranged  as  in  division, 
since  sucli  is  the  order  which  the  terms  assume  in  squaring  any 
polynomial  root. 

2nd,  In  squaring  any  polynomial,  the  first  term  of  the  square  is 
found  to  be  the  square  of  the  first  term  in  the  root ;  hence,  in 
extracting  the  square  root,  the  square  root  of  the  first  term  in  the 
given  polynomial  is  the  first  term  in  the  root. 

3rd.  To  prove  the  process  of  finding  the  divisors  and  the  subse- 
quent terms  of  the  root,  we  observe  the  following  operations : 

(1)  (2) 

B.  {a  +  l  +  cy  =  [_{a  +  l))  +  cf  =  (a  +  &)^  +  [2(a  +  5)  +  c> 

(1)  (2)  (3) 

=  a'^  4- (2«  +  &)&  +  [3(fl^  +  5)  +  c]c.  * 

C.  (a  +  l  +  c  +  df  =  [{a  +  'bJrc)  +  dY  =  (a  +  h  +  cY 

(1)  (2) 

+  [2{a  +  d  +  c)  +  d:]d  =  a^  +  (2a  +  b)h 

(3)  (4) 

+  [2(a  +  h)  +  c]c  +  [2(a  +  b  +  c)+d]d. 

Hence  it  appears;  1st,  That  the  square  of  a  polynomial  (as  a  +  h, 
a  +  i  +  c,  or  a  +  h  +  c  +  d)  is  made  up  of  as  many  parts  as  there  are 
terms  in  the  root ;  3nd,  That  the  first  is  the  square  of  the  first  term 
in  the  root ;  3rd,  That  the  second  part  is  Ticice  the  first  term  of  the 
root  (the  part  already  found),  +  the  second  term  of  the  root,  multi- 
plied by  the  second  term ;  4th,  That  any  one  of  these  parts,  as  the 
/7th,  is  Twice  that  portion  of  the  root  previously  found,  or  Twice 
the  n—1  preceding  terms  of  the  root,  +  the  wth  term  of  the  root, 
multiplied  by  this  last  or  «th  term. 

*  This  expaneion  is  made  by  treating  (a  +  b)  +  c  as  a  binomial,  giving  as  its 
square  (a  +  6) '  +  2  (a + 6)  c  +  c' .    Expanding'  and  factoring  we  have  the  form  in  B, 


EVOLUTIOlf.  159 

Scholium  I. — If  the  first  term  of  the  arranged  polynomial  is  not 
a  perfect  square,  the  root  can  not  be  extracted. 

Scholium  2.— If  at  anytime  no  term  of  the  remainder  can  be 
exactly  divided  by  the  first  term  of  the  trial  divisor,  the  root  can 
not  be  extracted. 

EXAMPLES. 
Ex.  1.  Extract  the  square  root  of  ^9a^a?i—30cu7^  +  2bx^ 

Model  Solution. 

OPERATION. 

16a* 


Sa'^—Sax !  —  24«»«  +  49aV 
-24a'a!+   9«V 


Sa^-6ax-\-5x^  \  40aV-30aa;«  +  25iB* 
40aV-30qa;H25a^ 

Explanation. — If  this  polynomial  is  a  perfect  square,  the  term 
containing  the  highest  power  of  a,  or  of  «,  is  the  square  of  the  first 
term  in  the  root.  Hence  I  place  16a*  first  in  the  arrangement. 
(25x*  would  do  as  well.)  And,  as  the  terms  arising  from  squaring 
a  polynomial  (e.  g.  the  root  of  this  given  number),  are  arranged 
according  to  the  leading  letter  of  the  root,  I  arrange  the  whole 
polynomial  according  to  the  powers  of  a,  as  this  will  be  the  leading 
letter  of  the  root,  when  I  put  16a*  fii*st  in  the  power. 

Having  arranged  the  given  number,  I  know  that  the  square  root 
of  the  first  term,  16a*,  or  4a''  is  the  first  term  of  the  root,  since  in 
squaring  any  polynomial  (e.  g.,  the  root  sought),  the  first  term  in 
the  square  is  the  square  of  the  first  term  in  the  root. 

Now,  having  removed  by  subtraction  the  square  of  the  first  term 
of  the  root,  I  double  the  root  already  found,  obtaining  8a'  for  a 
trial  divisor.  This  is  the  trial  divisor,  since  the  second  part  of  any 
square  (the  square  of  the  first  term  of  the  root  being  called  the  first 
part)  is  twice  the  root  already  found  +  the  next  term  of  the  root, 
multiplied  by  this  next  term.  I  now  find  that  the  trial  divisor  is 
contained  in  the  first  term  of  the  remainder  —Sax  times,  which  is, 
therefore,  the  second  term  of  the  root.  But  the  true  divisor  is  twice 
the  root  previously  found  plus  this  last  term;  hence  I  add  —'da.x 


160  POWERS    AND    ROOTS. 

to  the  trial  divisor  and  have  Sa"—Sax  as  the  trm  divisor.  Multi- 
plying this  true  divisor  by  the  last  term  of  the  root  I  obtain 
— 24a^«  +  9aV,  which  is  the  second  part  of  the  given  power. 

Subtracting  this  second  part  of  the  power,  the  third  part  is  twice 
the  root  already  found  +  the  next  term  of  the  root,  multiplied  by 
this  next  term.  In  this  case,  therefore,  the  new  trial  divisor  is  Sa'^ 
—Qax.  [Proceed  just  as  in  the  last  paragraph,  and  complete  the 
explanation.] 

Finally  4a^  — 3aa;  +  5«^  is  the  required  root,  for,  indeed,  I  have 
actually  squared  it  and  subtracted  this  square  from  the  given 
number  and  found  no  remainder.  [The  pupil  should  notice 
that  the  sum  of  the  several  subtrahends  is  the  square  of  the 
root.] 

2.  Extract  the  square  root  of  Sx-{-^-j-x*-{-^a:^-\-Sx^ 

Boot,  x^-\-2x  +  2. 

3.  Extract  the  square  root  of  25«V— 12«x3  +  16«4  +  4a:* 
—24:a^x.  Root,  2x^—dax-\-4:aK 

4.  Extract  the  square  root  of  x^—6aa^-^16a^x^—20a^x^ 
-^16a^x^—6a^x  +  a^. 

X?  4 

5.  Extract  the  square  root  of  x^—7?-\-j  -f-4rr— 2-f  -^ 

Root,  x^-^+-' 
2     X 


6.  Extract  the  square  root  of  9x—2^x^y^-{-12x^-\-16y'^ 
—  16yt_f-4.  Root,  Sx^—4:yi-{-2, 

7.  Extract  the  square  root  of   9a-^-{-12a-^i^—ea  +  4:b' 
—^a^^-\-a\  .  Root,  3a-^-\-2b^—a^. 

Query.— In  arranging  the  last  with  respect  to  a,  why  should  ih* 
come  before  —  6a  ? 


EVOLUTION.  161 

20S.  Scholium.— Since  the  square  root  of  a  quantity  is  either 
-f  or  — ,  all  the  signs  in  the  above  roots  may  be  changed,  and  they 
will  still  be  the  roots  of  the  same  polynomials.  Thus,  in  the  3rd 
Example,  if  we  call  the  square  rootof  4aj*,  minus  2x',  {^2x*,)  which 
it  is,  as  well  as  +2x^,  and  then  continue  the  work  as  before,  we  get 
for  the  root  '-2x^  +  3aa-4a\ 


209,  Prob.  3.— To  extract  the  Square  Root  of  a  Deci- 
mal Number  either  exactly  op  approximately. 

Rule.—/.  Separate  the  numhers  into  periods  by 
placing  a  marh  over  units  and  over  each  alternate 
figure  therefrom,  calling  the  marked  figure  with  the 
one  at  its  left,  if  any,  a  pei^iod.  The  number  of 
figures  in  the  root  is  equal  to  the  number  of  periods 
thus  formed, 

II,  Take  the  square  root  of  the  greatest  square  in 
the  left  hand  period,  and  write  it  as  the  highest  order 
in  the  root.  Subtract  the  square  of  this  figure  from 
the  period  used,  and  to  the  rem^ainder  annex  the 
next  period  for  a  new  dividend. 

HI.  Double  the  root  already  found  for  a  Trial 
Divisor,  by  which  divide  the  new  dividend,  rejecting 
in  the  trial  the  right  hand  figure  of  this  dividend. 
The  quotient  is  the  next  figure  in  the  root  or  a  greater 
one.  To  obtain  the  True  Divisor  annex  to  the  Trial 
Divisor  the  last  root  figure.  Multiply  this  True 
Divisor  by  the  last  root  figure,  subtract  the  product 
from  the  last  new  dividend,  and  to  the  remainder 
annex  the  next  period  of  the  given  number  for  another 
new  dividend. 

IV.  Double  the  root  already  found  for  a  new  Trial 
Divisor,  and  repeat  the  processes  given  in  the  3rd 
paragraph  till  all  the  periods  are  brought  down.  If 
the  number  is  a  perfect  square,  the  last  remainder  is 


162  POWERS    AN^D    ROOTS. 

zero.  If  not,  annex  periods  of  two  O's  each,  and  covj- 
tinue  the  process  till  the  required  degree  of  accuracy 
is  attained.  All  the  root  figures  arising  from  decimal 
fractional  periods  are  decimal  fractions. 

Scholium  I. — In  separating  decimal  fractions,  or  the  fractional 
part  of  mixed  numbers  into  periods,  make  full  periods  of  two  figures 
each,  annexing  a  0  if  necessary. 

Scholium  2. — If  at  any  time  the  Trial  Divisor  is  not  contained  in 
the  dividend  to  be  used,  annex  a  0  to  the  root  and  also  to  the  Trial 
Divisor,  and  then  bring  down  the  next  period  and  divide. 

Scholium  3. — When  the  work  does  not  terminate  with  the  last 
period  of  significant  figures  it  will  not  terminate  at  all,  and  the 
given  number  is  a  surd.  This  is  evident,  since  the  process  makes 
the  unit's  figure  in  each  subtrahend,  the  square  of  the  last  figure  in 
the  root,  but  no  figure  squared  gives  0  in  unit's  place.  The  process 
can,  however,  be  carried  to  any  given  degree  of  accuracy. 

Demonstration. — 1st.  That  this  method  of  pointing  gives  the 
number  of  places  in  the  root,  is  made  evident  by  squaring  a  few 
numbers.  Thus  the  square  of  1  is  1,  and  the  square  of  10  is  100, 
hence  the  squares  of  all  numbers  between  1  and  10  have  1  or  3 
figures ;  that  is,  Twice  as  unany  figures  as  the  root,  or  one  less  than  twice 
as  many.  Again,  the  square  of  100  is  10000 ;  hence  the  square  of 
all  numbers  between  10  and  100  have  3  or  4  figures;  that  is,  Twice 
as  many  as  there  are  in  the  root,  or  one  less.  In  like  manner  it  is 
readily  seen  that  the  square  of'  any  number  consists  of  Ticice  as 
many  figures  as  the  root,  or  one  less.  Hence  the  method  of  pointing 
indicates  the  number  of  figures  of  which  the  root  consists. 

In  the  case  of  decimal  fractions,  since  the  number  of  decimals  in  a 
product  equals  the  number  in  both  the  factors,  there  are  always 
twice  as  many  decimals  in  the  square  as  in  the  root.  Hence  if  the 
number  of  decimal  places  in  the  given  number  is  odd,  they  are  to  be 
made  even  by  annexing  0. 

2nd.  That  the  greatest  square  in  the  left  hand  period  is  the  square 
of  the  highest  order  in  the  root,  appears  from  the  facts  that  the 
square  of  any  number  oi  units  between  1  and  9  falls  in  the  first  right 
hand  period,  the  square  of  any  number  of  tens  between  1  and  9  falls 
in  the  second  period,  the  square  of  any  number  of  hundredshetwe&ii 
1  and  9  falls  in  the  third  period,  etc.  Moreover,  though  the  left 
hand  period  usually  contains  more  than  the  square  of  the  highest 


EVOLUTION.  163 

order  in  the  root,  it  can  not  contain  the  square  of  a  unit  more  of  that 
order,  since  all  the  figures  that  can  follow  this  highest  order  in  the 
root  can  not  make  another  unit  of  this  order.  Thus  the  square 
of  3999,  can  not  be  as  great  as  the  square  of  4000 ;  but  the  square 
of  4000  gives  16  in  the  highest  period  of  the  power,  hence  the  square 
of  3999,  must  give  less  than  16  in  that  period, 

3rd.  To  prove  the  method  of  finding  and  using  the  Trial  Divisor, 
suppose,  in  any  given  case,  the  pointing  shows  that  the  root  con- 
sists of  4  figures.  Represent  the  thousands  of  the  root  by  T,  the 
hundreds  by  A,  the  tens  by  t,  and  the  units  by  v.  The  number 
itself  will  be 

(T  -^h-ht  +  uy  =-.  r  +  {2T+h)h  +  [2(T+?i)  +  t]i  +  [2{T-\-h-{-t) 
+  u]u.     (207,  3rd  paragraph  in  the  Dem.) 

Whence  having  found  and  removed  the  square  of  the  thousands, 
7^,  the  next  part  of  the  power  is  {2T+h)h.  But  as  the  lowest 
order  of  this  is  hundreds  multiplied  by  hundreds,  or  10,000'8  we 
need  not  bring  down  anything  below  10,000's,  or  the  next  period, 
for  none  of  this  part  is  contained  in  the  lower  periods.  Again,  for 
trial,  considering  this  part  as  2Txh,  the  product  contains  nothing 
lower  than  hundred  thousands ;  hence  the  ten  thousands  figure  is  to 
be  omitted  in  the  trial  division.  But  the  true  divisor  is  2T+h\ 
hence  the  root  figure  is  annexed  to  the  trial  divisor,  or  really  added 
to  it,  regarding  the  k'cal  values. 

4th.  It  is  evident  that  this  process  is  merely  repeated  as  we 
increase  the  number  of  figures  in  the  root ;  and  as  the  law  of  nota- 
tion is  the  same  when  we  pass  the  decimal  point  into  a  fraction,  no 
special  exemplification  is  needed  in  such  a  case. 

EXAMPLES. 
Ex.  1.  Extract  the  square  root  of  7284601. 

Model  Solution. 
Operation.  728460i  (  2699 

46  )  328 
276 


529  )  5246 
4761 


5889  )  48501 
48501 


164  POWERS    AND    ROOTS. 

Explanation. — As  the  highest  order  in  this  number  is  millions,  the 
highest  order  in  the  root  is  thousands,  for  the  square  of  thousands  is 
millions  or  ten  millions,  and  the  square  of  anything  above  thousands 
is  more  than  ten  millions.  The  root,  therefore,  consists  of  thousands, 
hundreds,  tens,  and  units,  as  indicated  by  the  pointing. 

The  square  of  the  thousands  figure  in  the  root  evidently  must 
b3  sought  in  the  7  millions,  or  the  left  hand  period.  The  greatest 
square  in  this  being  4,  the  square  root  of  which  is  3,  the  thousands 
figure  of  the  root  is  2,  which  I  therefore  place  in  the  root. 

Now  the  root  may  be  represented  by  T+h  +  t  +  u,  T  standing 
for  the  thousands,  h  for  the  hundreds,  t  for  the  tens,  and  u  for  the 
units.     Hence  the  number  itself  will  be  represented  by  {T+h-\-t 

Having  removed  the  T'^  (2  thousands  squared)  from  the  number, 
the  second  part  is  {2T+h)K  the  lowest  order  in  which  is  hundreds 
multiplied  by  hundreds,  which  gives  ten  thousands;  wherefore  I 
bring  down  no  order  lower  than  ten  thousands,  or  simply  the  next 
period.  As  this  new  dividend  contains  {2T+?i)  h,  I  yvi\],fora  trials 
consider  it  as  simply  containing  {2T)xh,  and  as  hundreds  into 
thousands  produce  only  orders  aibove  ten  thousands,  I  may  omit  the 
8  which  is  ten  thousands,  in  making  this  trial.  Using  2jr  or  4 
(thousands)  for  the  trial  divisor,  I  find  it  contained  in  32  (hundred 
thousands)  8  times.  But,  as  the  trial  divisor  ii  too  small  by  this 
new  figure,  it  is  evident  that  adding  it,  thus  making  the  divisor  48, 
it  will  not  be  contained  8  times.  Neither  will  it  be  contained  7  times. 
Thus  I  find  6  to  be  the  next  figure  in  the  root,  and  the  true  divisor 
2T+h,  to  be  46  {i.  e.,  4  thousands  and  6  hundreds).  Multiplying 
46  by  6,  and  thus  forming  the  part  [2T+h]  /i,  I  find  it  to  be  276  (ten 
thousands),  which  subtracted  leaves  52  (ten  thousands). 

Again,  these  two  parts,  viz.,  T'-  and  {2T+h)h  having  been  re- 
moved, the  next  part  of  the  power  is  [2  (T+h)  +  t]  t,  or  [52  hundreds 
+  t]t.  As  the  lowest  order  of  this  part  is  tens  squared,  I  need  bring 
down  nothing  below  hundreds,  or  the  next  period.  The  pupil  can 
now  fill  out  the  demonstration  as  in  the  preceding  paragraph. 

2.  Extract  the  square  root  of  7225.  Boot,  85.     * 


3  to  7.   Show  that  V9801  =  99,  V47089  =  217, 
\/553536  =  744,  a/43046721  =:  6561,  ^5764801  =  2401. 

8  to  11.  Show  that  V^  =  .7071+,  VS  ==  1.73  +  , 
^50  =  7.071 +,  V5  =  2.236  +  . 


EVOLUTlOK.  165 

210.  Cor. — In  extracting  thr  roots  of  common  fractions, 
if  the  numerator  and  denominator  are  perfect  squares,  extract 
the  root  of  each  separately ;  if  they  are  not,  it  is  usually  best 
to  convert  the  fraction  into  a  decimal  and  then  extract  its  root. 

12  to  15.  ^/%^'^,  y/l^-n^+,  \/i--577+. 


V 


I  =  .8164+. 


211.  Prob.  4.— To  extract  the  Cube  Root  of  a  Poly- 
nomial. 

Rule. — /.  AiTange  the  polynomial  with  reference  to 
one  of  its  letters,  as  for  division. 

II.  Extract  the  cube  root  of  the  first  left  hand  term. 
Tliis  root  is  the  first  term  of  the  required  root. 
Subtract  the  cube  of  this  term  of  the  root  from  the 
polynomial. 

III.  Tahe  three  times  the  square  of  the  root  already 
found  for  a  Trial  Divisor.  By  this  trial  divisor 
diiride  the  first  term  of  the  remainder  of  the  polyno- 
mial, and  write  the  quotient  as  the  second  term  of  the 
root. 

IV.  Complete  the  divisor  by  adding  to  the  trial 
divisor  3  times  this  last  term  multiplied  by  the  part 
of  the  root  previously  found,  and  also  the  square  of 
the  last  term  found.  Multiply  the  ti^ue  divisor  thus 
formed  by  the  last  term  in  the  root,  and  subtract  the 
product  frowj  the  last  remainder,  bringing  down  such 
terms  as  may  be  necessary. 

V.  Repeat  the  process  of  forming  Trial  Divisors, 
dividing,  completing  the  divisor,  multiplying  and 
subtracting,  till  the  polynomial  is  exhausted,  or  until 
there  is  no  term  of  it  remaining  which  can  be  exactly 
divided  by  the  first  term  of  the  Trial  Divisor, 


166  POWERS    AND    HOOTS. 

Demonstration. — 1st.  The  polynomial  is  arranged  as  in  division, 
since  this  is  the  order  which  the  terms  assume  in  cubing  any  poly- 
nomial, similarly  arranged. 

3nd.  In  cubing  any  polynomial,  the  first  term  of  the  cube  is  found 
to  be  the  cube  of  the  first  term  of  the  root;  hence,  in  extracting  the 
cube  root,  the  cube  root  of  this  term  is  the  first  term  of  the  root. 

3d.  To  prove  the  process  of  finding  the  divisors  and  subsequent 
terms  of  the  root,  we  observe  the  following  operations  : 

(1)  (2) 

A.  (a  +  dy  =  a^  +  3a«5  +  3a5'  +  &'  =  «'  +  [3a'  +  Sai + ¥]  6. 

B.  {a  +  h  +  cy  =  [(a  +  l>)  +  cY 

(1)  (2)  (3) 

=  a^  +  [Ha''  +  dab  +  l*'\J)  +  [d  (a  +  hy  +  3  (a  +  h)  c  +  c^]  c. 

C.  (a+h+c+dy  =  [ia+d  +  c)  +  d]' 

=  (a  +  I}^cy  +  [d(a  +  l>  +  cy  +  3(a  +  l}  +  c)d  +  d'']d 

(1)  (2)  (3) 

=  a»  +  [3a'  +  3a&  +  6'']&  +  [3(a  +  &f +  3(a  +  5)c  +  c''](j 

(4) 
+  [d(a  +  l  +  cy  +  d(a  +  l+c)d  +  d']d. 

Hence  it  appears ;  1st,  That  the  cube  of  a  polynomial  is  made  up 
of  as  many  parts  as  there  are  terms  in  the  root ;  2nd,  that  the  Jirst 
part  is  the  cube  of  the  first  term  of  the  root ;  3d,  That  the  second 
part  is  three  times  the  square  of  the  Jirst  term  of  the  root  +  3  times  the 
first  term  into  the  second  term  +  the  square  of  the  second  term,  multi- 
plied lyy  the  second  term  of  the  root ;  4th,  That  any  one  of  the  parts  of 
the  power,  as  the  nth,  is  Three  times  the  square  of  the  n—l  preceding 
terms  of  the  root,  +  3  times  the  product  of  these  terms  into  the  next,  or 
jith  term.,  +  tJie  square  of  this  last  a?'  nth  term,  all  these  terms  heing 
multiplied  by  the  last,  or  nth  term  of  the  root. 

Finally,  it  is  evident  that,  if  the  work  does  not  terminate  by  this 
process  when  the  letter  of  arrangement  disappears  from  the  remain- 
der, it  can  never  tern#nate,  since  the  divisor  always  contains  this 
letter. 

Scholium  I. — If  the  first  term  of  the  arranged  polynomial  is  not 
a  perfect  cube  the  root  cannot  be  (ixtracted. 

Schoh'um  2. — If  at  any  time  no  term  of  the  remainder  is  exactly 
divisible  by  the  first  term  of  the  trial  divisor,  the  root  can  not  be 
extracted. 


EVOLUTIOK. 


167 


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168  POWERS    AND    ROOTS. 

Explanation. — 1st.  I  arrange  this  polynomial  with  reference  to«, 
and  thus  see  at  once  the  first  two  terms.  But  the  terms  36aV  and 
27a*6c'^a!  are  of  the  same  degree  with  respect  to  a,  and  hence  to 
determine  which  is  to  have  the  precedence,  I  notice  that  the  first 
term  in  the  root  will  be  Sa^c,  and  as  the  second  term  of  the  poly- 
nomial divided  by  3  times  the  square  of  this  gives  the  second  term 
of  the  root,  I  observe  that  the  terms  containing  a  and  c  are  all  to 
have  precedence  over  those  containing  &  and  x.  Uence  I  write 
dQa*c—8a^  next.  The  remaining  terms  I  arrange,  giving  a  the 
precedence  and  noticing  that  as  x  will  be  in  the  last  term  of  the 
root,  its  higher  powers  will  stand  last. 

2nd.  As  27a"c'  is  the  cube  of  the  fii*8t  term  of  the  root,  that  term 
is  ^\  which  I  consequently  place  in  the  root,  and  subtract  the 
term  27 a^c^  from  the  polynomial. 

3rd.  As  the  second  part  of  a  cube  of  a  polynomial  is  3  times  the 
square  of  the  first  term  of  the  root,  plus  other  terms,  into  the 
second  term  of  the  root,  I  take  3  times  the  square  of  this  first  term 
of  the  root  or  27a*c',  for  a  trial  divisor.  Dividing,  I  find  the  second 
term  of  the  root  to  be  —2a.  But  the  True  Bwiaor,  or  leading 
factor  in  this  second  part  of  the  power,  is  3  times  the  square  of  the 
former  part  of  the  root,  +  three  times  that  part  into  the  last  term 
found,  +  the  square  of  this  term.  Hence  I  add  3  times  Sa^c  multi- 
plied by  —2a,  and  —2a  squared,  to  complete  the  divisor.  Having 
completed  it,  I  multiply  it  by  the  last  term  of  the  root  found,  —2a, 
and  thus  form  the  second  part  of  the  power  of  the  root,  which  I 
subtract  from  the  given  polynomial. 

4th.  The  explanations  of  the  next  and  succeeding  steps,  when 
there  are  more,  are  identical  with  the  last,  and  can  be  supplied  by 
the  student. 


2.  Extract  the  cube  root  of  a^—Sb^  +  12aIyi—{ 

Root,  a— 21). 

3.  Extract  the  cube  root  of  boi?—l—doiP  +  x^^Zx. 

Root,  x^—x—l. 

4.  Extract  the  cube  root  of  66a^  +  1  —  Q3a^  —  ^x  +  ^ofi 
—mx^+^ZxK  Root,  2a;2_3a:+l. 


EVOLUTION. 


169 


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170  POWERS    AJ^D    ROOTS. 

2 12,  Prob.  5.— To  extract  the  cube  poot  of  a  decimal 
number,  either  exactly  op  apppoximately. 

Rule. — /.  Separate  the  nitmber  into  periods  hy  pla- 
cing a  marh  over  units  and  over  each  third  figure 
therefrom,  calling  the  figure  marked,  together  luitli 
the  two  at  its  left  (if  there  are  so  many),  a  period. 
The  number  of  places  in  the  root  is  the  same  as  the 
number  of  periods. 

II.  Take  the  cube  root  of  the  greatest  cube  in  the 
left  hand  period,  and  write  it  as  the  highest  order  in 
the  root.  Subtract  the  cube  of  this  figure  from  the 
period  used,  a,nd  to  the  remainder  annex  the  next 
period  for  a  new  dividend. 

III.  Take  three  times  the  square  of  the  root  already 
found,  regarding  it  as  tens,  for  a  trial  divisor,  by 
which  divide  the  new  dividend.  The  quotient  is  the 
next  figure  in  the  root  (or  a  greater  one).  To  obtain 
the  True  Divisor  add  to  the  trial  divisor  3  times  the 
product  of  the  last  root  figure  by  the  preceding  part  of 
the  root,  regarded  as  tens,  and  also  the  square  of  the 
last  figure  in  the  root.  Multiply  this  true  divisor  by 
the  last  root  figure  and  subtract  the  product  from  the 
last  new  dividend,  and  bring  down  the  next  period. 

IV.  Repeat  this  process  till  all  the  periods  have  been 
brought  down.  If  the  number  is  a  perfect  cube  the 
remainder  is  zero.  If  not,  annex  periods  of  3  zeros 
each  and  continue  the  operation  till  the  required 
degree  of  accuracy  is  attained,  marking  the  figures 
thus  obtained  as  a  decimal. 

Scholium  I. — In  pointing  off  decimal  fractions,  or  the  fractional 
part  of  mixed  numbers,  make  full  periods  of  three  figures  each, 
annexing  O's  if  necessary. 

Scholium  2.— If  at  anytime  the  trial  divisor  is  not  contained 
in  the  dividend  to  be  used  according  to  the  3rd  paragraph  in  the 


EVOLUTIOK.  171 

rule,  annex  a  0  to  the  root  and  also  two  zeros  to  the  trial  divisor, 
bring  down  the  next  period,  and  then  divide. 

Scholium  3. — When  the  work  does  not  terminate  with  the  last 
period  of  significant  figures  it  will  not  terminate  at  all,  and  the 
number  is  a  surd.  This  is  evident  since  the  right  hand  figure  in 
any  subtrahend  arises  from  cubing  the  corresponding  digit  in  the 
root,  and  the  cube  of  no  digit  produces  0  in  unit's  place. 

Demonstration. — 1st.  That  this  method  of  pointing  gives  the 
number  of  figures  in  the  root  is  made  evident  by  cubing  a  few  num- 
bers. Thus  the  cube  of  1  is  1,  and  of  10  is  1000  ;  hence  the  cubes 
of  all  numbers  between  1  and  10  have  1,  2,  or  3  (cannot  have  4) 
figures.  The  cube  of  100  is  1,000,000;  hence  the  cube  of  numbers 
between  10  and  100  have  4,  5,  or  6  figures,  but  can  not  have  7.  Again, 
the  cube  of  1000  is  1,000,000,000;  hence  the  cube  of  any  number 
between  100  and  1000,  i.  e.,  of  any  number  represented  by  3  figures, 
contains  9,  or  one  or  two  less  than  9  figures.  In  like  manner  it  appears 
that  the  cube  of  any  integral  number  contains  either  three  times  as 
many  Jig  ures  as  the  root,  or  one  or  two  leas.  In  the  multiplication  of 
decimal  fractions  the  number  of  fractional  places  in  the  product  is 
equal  to  the  number  in  both  or  all  the  factors  used,  hence  the  frac- 
tional part  of  any  cube  must  have  three  times  as  many  figures  as 
the  root. 

2nd.  That  the  greatest  cube  in  the  left  hand  period  is  the  cube 
of  the  highest  order  in  the  root,  appears  from  the  facts  that  the  cube 
of  any  number  of  units  between  1  and  9  falls  in  the  1st  period  ;  the 
cube  of  any  number  of  tens  between  1  and  9,  falls  in  the  second ;  of 
any  number  of  hundreds,  in  the  3rd,  etc.  Moreover,  though  the  left 
hand  period  usually  contains  more  than  the  cube  of  the  digit  in  the 
highest  order  in  the  root,  it  can  not  contain  the  cube  of  a  unit  more 
of  that  order,  since  all  the  figures  that  can  follow  this  highest  order 
in  the  root  can  not  make  another  unit  of  that  order.  Thus  the  cube 
of  3999  can  not  be  as  great  as  the  cube  of  4000.  But  the  cube  of 
4000  gives  64  in  the  highest  period.  Hence  the  cube  of  3999  must 
give  less  than  64  in  that  period. 

3rd.  In  any  given  case,  suppose  the  pointing  shows  that  the  root 
consists  of  thousands,  hundreds,  tens,  and  units.  Represent  the 
thousands  by  T,  the  hundreds  by  A,  the  tens  by  t,  and  the  units  by 
u.  Then  the  number  is  (T-{-h  +  t-{-uy  =  T^-\-[ST''  +  STh-\-h^]h 
^[diT+hy  +d{T-{-h)t-\-t']t-]-[^{T+h-{-ty-\-S(T-\-h  +  t)u-\-2r]u. 
But  having  removed  the  cube  of  thotliousands,  T'^  the  next  part  of 
the  power  is  [37"  +  3rA  +  A'J/i.     No  part  of  this  can  fall  in  either 


172  POWERS    AND    ROOTS. 

of  the  two  lowest  periods  of  the  power,  since  its  lowest  order  arises 
from  h^  which  is  1,000,000,  at  least.  Hence  we  need  only  bring 
do^\Ti  one  period.  For  a  tiial,  we  consider  this  part  as  37"^  x  h,  and 
hence  the  Trial  Divisor  is  3T^,  or  '^timesthe  square  of  the  root  already 
found.  Again,  regarding  this  thousands  figure  as  tens,  n\akes  the 
T,  which  squared  and  multiplied  by  the  next  figure  of  the  root 
which  is  also  hundreds,  give  millions,  the  same  order  as  the  new 
dividend.  But  the  True  Divisor  is  ^T'^ -\-^Th  +  h^\  hence  we  add  to 
ZT\  '6Th  +  ¥  ;  i.  e.,  3  times  the  root  previously  found  multiplied  ly  the 
last  figure^  and  the  square  of  this  last  figure.  In  making  this  correc- 
tion we  are  to  remember  to  call  the  thousands  so  many  tens^  which 
reduces  it  to  hundreds,  the  order  of  the  root  which  we  are  seeking ; 
whence  the  correction  becomes  the  square  of  hundreds,  or  of  the 
same  order  as  the  trial  divisor,  and  can  be  added  to  it. 

4th.  It  is  evident  that  this  process  is  merely  repeated,  as  we  pro- 
ceed to  obtain  other  figures  in  the  root ;  and,  as  the  law  of  notation 
is  the  same  as  we  pass  the  decimal  point,  no  special  exemplification 
is  needed  in  that  case. 

EXAMPLES. 
Ex.  1.  Extract  the  cube  root  of  99252847. 
Model  Solution.— Operation. 


h  t  u 
99252847  I  4  6  3 
64  ' 


IHa?  Divisor  3  (40)^  =  4800 
Corrections  I  ^(,^?)«=  ^3^^ 
True  Divisor  . 5556 


35252 


33336 


Trial  Divisor  3  (460)^  =634800 
4140 
9 
True  Divisor 638949 


Corrections  j^(^3^^^>^^ 


1916847 
1916847 


Explanation. — As  the  highest  order  in  this  number  is  ten  7nillion8, 
the  highest  order  in  the  root  is  hundreds,  since  the  cube  of  a  hun- 
dreds figure  falls  in  millions  period,  while  the  cube  of  thousands 
falls  in  billions.  Moreover,  the  cube  of  the  hundreds  figure  is  the 
greatest  cube  contained  in  99,  /.  e.  64,  the  root  of  which  is  4,  which 
is,  therefore,  the  hundreds  of  the  root.  That  the  hundreds  figure 
is  not  greater  than  4,  is  evident,  since  the  cube  of  5  hundreds  is 
greater  than  the  given  number. 


EVOLUTION.  173 

Therefore  the  cube  root  of  the  given  number  is  h-\-t  +  u,  and  the 
number  itself  is  (h  +  t  +  uy  =  h'-\-  [3  h'  +  dht  -^.t']t  +  [B{h  +  ty-\-  (3A 
+  t)u+u^]u.  But  having  removed  the  h^  by  subtracting  the  64 
(millions),  the  next  part  of  the  power  is  [3A*  +  3A< +  «"]«.  Now  the 
lowest  order  of  this  is  t\  or  the  cube  of  tens,  whicl)  cannot  fall 
below  thousands,  so  that  I  need  only  bring  down  thousands  period, 
/.  €.  the  next  lower.  For  a  Trial  I  now  consider  this  part  of  the 
power  (35352)  as  3^"  x  <,  or  3[40J'^  xt.  I  reduce  the  4  hundreds  to 
the  same  order  as  the  root  figure  which  I  am  seeking,  so  that  the 
product  of  its  square  by  this  root  figure  shall  be  of  the  same  order 
as  the  new  dividend.  Therefore,  reducing  the  4  hundreds  to  tens 
it  becomes  40,  whence  3(40)^  =  4800,  which  being  hundreds,  goes 
into  the  new  dividend,  which  is  thousands,  tens  times.  This  trial 
divisor  is  really  contained  in  the  dividend  7  {tens)  times,  but  as  the 
corrections  to  be  made  upon  it  for  the  true  divisor  are  so  great,  the 
true  divisor  is  contained  but  6  times,  as  I  find  by  trying  7  for  the 
tens  of  the  root.  Having  thus  found  6  to  be  the  tens  of  the  root,  I 
correct  my  trial  divisor,  which  by  the  formula  is  3^"  +  3A  x  ^  +  ^'^,  by 
adiling  3A  x  t  or  3(40)  x  6,  and  t"  or  Q^,  and  find  the  true  divisor  to 
be  5556  (hundreds).  This  multiplied  by  the  6  (tens)  gives  the 
second  part  of  the  power,  ^.  e.  (3A*  +  3/*<  +  0^  =  33336  (thousands), 
which  I  therefore  subtract  from  the  given  number. 

[The  next  step  is  exactly  like  the  last,  and  the  pupil  can  supply 
the  demonstration.] 

2.  What  is  the  cube  root  of  74088?  Ans.,  42. 

3.  What  is  the  cube  root  of  12326391  ?         Ans.,  231. 

4.  AVhat  is  the  cube  root  of  122097755681  ?  Ans.,  4961. 
6.  What  is  the  cube  root  of  2936.493568?  Ans.,  14.32. 

6.  What  is  the  cube  root  of  12.5  ?  Ans.,  2.321  nearly. 

7.  What  is  the  cuhe  root  of  .64  ?  Ans.,  .8617  +  . 

8.  What  is  the  cube  root  of  .08  ?  Ans,,  .43084-. 

9.  What  is  the  cube  root  of  .008  ?  Ans.,  .2. 
10  to  12.   Show    that    \^  =  1.2599 +;    \/5  =  1.7099 +  ; 

V^  =  2.08008  +  . 
13  to  15.  Show  that  ^|  =  . 87+;  \^Wh  =  iiy  <^^m 


174  POWERS    AND    ROOTS. 

[Note. — With  regard  to  the  various  shorter  methods  for  extract- 
ing roots,  the  various  methods  of  approximation  and  the  like,  no 
mathematician  thinks  of  using  them,  or  even  those  here  given,  but 
resorts  at  once  to  the  table  of  logarithms.  It  is  better  that  the 
student  should  spend  his  time  in  becoming  perfectly  familiar  with 
the  demonstration  of  a  single  method,  than  to  cumber  the  memory 
with  a  multiplicity  of  processes  which  he  will  not  remember,  and 
which  if  we  were  to  remember  he  would  never  use.] 


FOR  REVIEW  OR  ADVANCED  COURSE. 

213,  Prob.  6. — To  extract  roots  whose  indices  are 
composed  of  factors  2  and  3. 

Solution.— To  extract  the  4th  root,  extract  the  square  root  of 
the  square  root.  Since  the  4th  root  is  one  of  the  4  equal  factors 
into  which  a  number  is  conceived  to  be  resolved,  if  we  first  resolve 
a  number  into  2  equal  factors  (that  is,  extract  the  square  root)  and 
then  resolve  one  of  these  factors  into  3  equal  factors  (that  is,  extract 
its  square  root)  one  of  the  last  factors  is  one  of  the  4  equal  factors 
which  compose  the  original  number,  and  hence  the  4th  root.  In 
like  manner  the  6th  root  is  the  cube  root  of  the  square  root,  etc. 

EXAMPLES. 

Ex.  1.  What  is  the  4th  root  of  l^a^-maH-^^Ua^o^ 
—216«rr3^  81:^4?  Ans.,   ±(2a— 3^;). 

2.  What  is  the  6th  root  of  lbx^—%Qio^  +  ofi—Qo^-\-l  —  Qx 
+  15^:4?  Ans.y   ±{x—l), 

3.  What  is  the  6th  root  of  2985984? 

4.  What  is  the  8th  root  of  1679616  ? 


214.  Prob.  7.— To  extract  the  mX\\  (any)  root  of  a 
decimal  number. 

Solution. — Any  root  can  be  extracted  by  a  process  altogether 
similar  to  those  given  for  the  square  and  cube  roots,  or  by  a  simple 
inspecticm  of  the  corresponding  power  of  a  binomial.  Thus  to  extract 
the  fifth  root,  point  off  by  placing  a  point  over  units  and  every 


EVOLUTION.  175 

fifth  figure  therefrom,  for  the  7th  root  over  every  7th  figure,  etc. 
Extract  the  required  root  of  the  largest  power  of  the  7//th  degree 
in  the  let\;  hand  period  for  the  first  figure  in  the  root.  Subtract, 
and  bring  down  the  next  period.  To  form  the  trial  and  true  divisors, 
and  hence  to  find  the  other  figures  of  the  root,  consider  the  corres- 
ponding power  of  a  binomial.  Thus  for  the  5th  root,  we  have 
(^-f  &)'*  =  a^  +  5a^&-f  lOa^ft'  -H  lOa'ft^  +  5aJ*  +  J"  =  a"^  +  [5a*  +  lOa^J 
+  load's' +  5a5'']&.  The  trial  divisor  is  5a*,  i.  e.  five  times  the  4th 
power  of  the  root  already  found  regarded  as  tens.  The  corrections 
are  10«'&4-10a'&^  +  5a&'',  regarding  a  as  the  root  already  found  and 
as  tens,  and  &  as  the  next  figure,  i.  e.  the  one  sought  by  the  trial. 

In  the  7th  root  the  trial  divisor  is  7a'',  and  the  corrections  are 
21a»&  +  35a*6»  +  35a='&^  +  21a^&*  +  laV  +  h\ 

But  in  these  cases,  and  much  more  in  the  case  of  higher  roots,  the 
trial  divisor  differs  so  much  from  the  true  divisor  that  the  process 
is  little  better  than  guess-work. 


216.  Prob.  8.— To  extract  the  iwth  root  of  a  polynomial. 

Rule. — /.  Having  arranged  the  polynomial  as  for 
division,  take  the  root  of  the  first  teimv,  for  the  first 
term  of  the  required  root. 

II.  Subtract  the  power  from  the  given  quantitij,  and 
divide  the  first  term,  of  the  remainder  hij  the  first  term 
of  the  root  involved  to  the  next  inferior  power,  and 
multiplied  by  the  index  of  the  given  power ;  the  quo- 
tient will  be  the  next  term  of  the  root. 

III.  Subtract  the  power  of  the  terms  already  found 
from  the  given  quantity,  and  using  the  same  divisor 
proceed  as  before. 

Demonstration.— This  rule  demonstrates  itself,  as  the  final  opera- 
tion consists  in  involving  the  root  to  the  required  degree. 

Scholium.— This  rule  may  also  be  used  for  decimal  numbers. 

EXAMPLES. 
Ex.  1.    Find   the   fourth   root  of   16a*-ma^x-]-216a^xi 

2.  Find  the  fifth  root  of  a:«  +  5a;*  -f-  10a:» + lOar^ + 5a;  + 1. 


176 


POWERS    AND    ROOTS. 


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REDUCTIONS. 

217,  Prob.  1.— To  simplify  a  radical  by  removing  a 
factor. 

Rule. — Resolve  the  number  under  the  radical  sign 
into  two  factors,  one  of  which  shall  he  a  perfect  power 
of  the  degree  of  the  radical.  Extract  the  required 
root  of  this  factor  and  place  it  before  the  radical  sign 
as  a  coefficient  to  the  other  factor  under  the  sign. 

Demonstration. — This  process  is  simply  an  application  of  Cor. 
Art.  205,  which  proves  that  the  product  of  the  roots  is  equal  to 
the  root  of  the  product.   Thus  V^Sa'^  =  A/l6a*&'  x  Zdb  =  Vl6«¥ 

EXAMPLES. 
Ex.  1. — Reduce  to  its  simplest  form  v^TSG^^c^. 
Model  Solution. 


Operation 

ExpFanation.  \/lS9a*W  indicates  that  189a^&V 
into  3  equal  factors. 
27aW  and  7ac\  one  of  which  is  a  perfect  cube.  Now  I  resolve  each 
of  these  two  factors  into  3  equal  factors,  and  taking  one  of  each 
three  I  multiply  them  together.  This  product  will  constitute  one 
of  the  three  equal  factors  of  the  given  number.    37a'&'  =  da^b  >'  Sah 


''is  to  be  resolved 
Therefore  I  first  separate  it  into  two  factors, 


I 


REDUCTIONS.  179 

X  Sab,  and  Inc^  =  v^7^  x  v^W  x  ^7a^.     Hence  'iab\/7^^  is 
one  of  the  3  equal  factors  of  189a*JV;  or  \/ma*b'7-  =  'Sab\/Ta^. 

2.  Reduce  V^1!a^^  to  its  simplest  form. 

Result i  Sax^V^cuT' 

3.  Reduce  VSOa^afi  to  its  simplest  form. 

4.  Reduce  A/3?5r*^^  to  its  simplest  form. 

Suggestion. — If  the  factor  of  the  decimal  number  under  the 
radical  is  not  aj^parent,  it  can  readily  be  found  by  a  few  trials. 
Thus  in  the  3rd,  we  could  try  the  square  of  2,  or  4 ;  and  then  the 
square  of  3,  or  9  ;  and  then  of  4,  or  16 ;  of  5,  or  25  ;  of  6,  or  36  ;  of  7, 
or  49  ;  of  which  we  should  find  16  to  be  the  greatest  square  factor. 
We  need  not  try  farther  than  49,  since  this  is  more  than  ^  of  80,  and 
no  larger  number  can  be  a  factor. 

5.  Simplify  VllSS^^^ 

Suggestion. — Try  the  square  numbers  from  4  upward  till  you 
find  the  required  factor.  But  a  little  judgment  will  save  labor. 
Thus,  we  need  not  try  4,  for  no  numl)er  multiplied  by  4  gives  a  3  in 
units'  place.  For  a  like  reason  we  would  try  9,  but  not  16,  25  or 
36.  Then  again  we  would  try  49,  but  not  64 ;  81,  but  not  100;  121, 
but  not  144.  Finally  169  meets  the  case,  and  we  have  ^/TlS3x^ 
=  Vl69urVx"7^  =  ^/WxY  x  \/7^  =  Ida^y^^Txy. 

6.  Simplify  V^H^aW. 

7.  Simplify  v^96a^9.  Result,  Ux^dcM. 

8.  Simplify  ^'fmj^y^^^.        Result,  Qa^y'^^  \/Tof, 

9.  Simplify  (11ll5a^-^I^^)i. 

Result,  7a3m^»  v^5a^,  or  Ha^P^l^'f- 

10.  Simplify  (352«tjio)i  Result,  2a*^^^. 

Scholium. — Of  course,  by  the  use  of  fractional  exponents,  all  the 
factors  of  such  monomials  may  be  written  separately,  as  in  (202). 
Thus,  the  result  in  Ex.  7,  may  be  written  (QQ)lai!^^,  or  2(3)Wla^^, 
or  2ax(3)^af.r^,  or,  by  taking  the  5th  root  of  the  product  instead 
of  the  product  of  the  5th  roots  of  the  last  3  factors,  2ax(Sa^x*)^,  as 
above.  The  method  of  this  rule  is  usually  applied  for  removing 
factors  which  can  be  expressed  without  fractional  exponents. 


180  CALCULUS    OF    RADICALS. 


11.  Simplify  "s/a^—ahc.  Result,  a  y/a—x. 


12.  Simplify  ^a^^^H^.  Result,  aby/a"^. 


13.  Simplify  V{a^—IP){a-\-b).   Result,  (a  +  h)  Va—b. 

14.  Simplify  3^50^-  Result,  Ux^V^. 

Suggestion. — When  the  radical  has  a  coefficient,  the  factor 
removed  from  under  the  radical  sign  is  to  be  multiplied  into  this 
coefficient. 


16.  Simplify  {x-\-y)  ^^a^—^x^y-^-xy^. 

Result,  (x^—y^)  Vx. 

16.  Simplify  xyVx^y^—x^y^  Result,  a^y^Vx—y. 

17.  Simplify  ^x^^y-^^+K  Result,  x^y^^^z, 

18.  Reduce  A/f  to  its  simplest  form.         Result,  ^a/6. 

Scholium. — A  surd  fraction  is  conceived  to  be  in  its  simplest 
form  when  the  smallest  possible  whole  nuinber  is  left  under  the 
radical  sign.  The  reason  for  this  is,  not  only  that  the  radical  factor 
is  thus  made  simpler,  but,  if  a  fraction  were  to  be  left  under  the  radical 
sign  the  question  would  arise,  What  fraction  ?  Certainly  not  the 
least  possible,  for  such  a  fraction  can  be  diminished  at  pleasure. 

Thus,  /I  =  /4  X  ^  =  2/gU  2/16  X  g^  =  8|/i.  etc.,  with- 

out  limit.     Perhaps,  if  a  fraction  is  to  be  left  under  a  radical  sign 
it  will  be  proper  to  consider  the  expression  as  simplest  when  the 

fraction  is  nearest  unity ;  whence  y  -  is  to  be  considered  as  simpler 
than  84/1. 

218*  Cor. — The  denominator  of  a  surd  fraction  can  always 
be  removed  from  under  a  radical  sign  by  multiplying  both  terms 
of  the  fraction  by  some  factor  which  will  make  the  denominator  a 
perfect  power  of  the  degree  required. 


REDUCTIONS.  181 


sim- 


19.  Reduce  W-,  y  ^,  W^-,  and  W-^-  to  their 

plest  forms. 

Scholium. — The  root  of  a  fraction  having  1  for  its  numerator  is 
equal  to  the  same  fraction  into  the  root  ot  the  next  lower  power  of 

the  denominator.     Thus,  |/i  =  1  ^7,  f"^  =  ^  V^49,  \/\  ^   \ 
V^343,  etc. 

20.  Reduce  \/  jk  to  its  simplest  form. 

21.  Reduce    r  to  its  simplest  form. 

Result,  — — -  Va^—a^. 
a-\-x 

22.  Reduce  ttx/  ^,  SVIS'  ^^^     V  W  ^^  ^^^^  ^^^' 
plest  forms. 

n 

Remits,  (not  in  order),  ^  V\bx,  t  a/6«^,  and  —  a/7. 


23.  Reduce  -^ — Th\/ io — ■ to  its  simplest  form. 

^2— J2y  62 


Result,  -r-, — TTv  v/«*. 


24.  Reduce  V/o  »  ^"^  V  4  ^^  *^^^^  simplest  forms. 


l..,...._,..........„ 

^t  fractions,  but  upon  integral  radicals  only  when  the  integer  has  a 
^^yactor  which  is  a  perfect  power  of  the  degree  of  the  radical. 

P  .     . 

219.  Prob.  2.— To  simplify  a  radical,  op  reduce  it  to 
its  lowest  terms,  when  the  index  is  a  composite  number, 
and  the  number  under  the  radical  sign  is  a  perfect  power  of 
the  degree  indicated  by  one  of  the  factors  of  the  index. 


18^  CALCULUS   OF   RADICALS. 

Rule. — Extract  that  root  of  the  number  which 
corresponds  to  one  of  the  factors  of  the  index,  and 
write  this  root  as  a  surd  of  the  degree  of  the  other 
factor  of  the  given  index. 

Demonstration.— The  wwth  root  is  one  of  the  mn  equal  factore 
of  a  number.  If,  now,  the  number  is  resolved  first  into  m  equal 
factors,  and  tlien  one  of  these  m  factors  is  again  resolved  into  n 
other  equal  factors,  one  of  the  latter  is  the  mni\\  root  of  the  number. 

Illustration. — The  4th  root  of  a  number  is  one  of  the  four  equal 
factors  of  that  number ;  if  we  resolve  the  number  into  2  equal  fac- 
tors, and  then  one  of  these  factors  into  2  other  equal  factors,  one  of 
the  latter  is  one  of  the  4  equal  factors  which  compose  the  given 
number. 

EXAMPLES. 

Ex.  1.  Reduce  \/mi^. 

Model  Solution. 

Operation.     'V^257?J^  =  Vy^SS^?  =  V^^'  =  abV^. 

Explanation. — The  4th  root  of  25a*6"  is  one  of  the  4  equal  factors  of 
it.  Hence  I  first  resolve  it  into  two  equal  factors,  one  of  which  is 
5a^5^  Then  I  resolve  5«^5^  into  two  equal  factors,  or  rather  indicate 
it,  as  the  operation  cannot  be  fully  performed,  and  have  \/5a^¥. 
/^ba^h^  is,  therefore,  one  of  the  4  equal  factors  of  25a^&".  But,  by 
the  last  problem,  \/^aFb^  =  a'b\/'5b.     Hence  V^S^^F  =  db\/^. 


2.  Reduce  ^27a^K  Result,  hVSab. 


Suggestions.   ^27a'b'  =  V\^27a'b'  =  \^3ab'  =  h^/S^. 

3.  Reduce  ^y—64:a^  Result,  2V^^. 

4.  Reduce  \/266a^^. 


5.  Reduce  VSln^mK 


6.  Reduce  V^—^^y-^-y^. 


220,  Prob.  3. — To  reduce  any  number  to  the  form  of  a 
radical  of  a  given  degree. 


REDUCTIONS.  183 

Rule. — Raise  the  nimibcr  to  a  power  of  the  same 
degree  as  the  radical,  and  place  this  power  under  the 
radical  sign  with  the  required  index,  or  indicate  the 
same  thing  by  a  fractional  exponent. 

Demonstration. — That  this  process  does  not  change  the  value  o£ 
the  expression  is  evident,  since  the  number  is  first  involved  to  a 
given  power,  and  then  the  corresponding  rpot  of  tliis  power  is 
indicated,  the  latter,  or  indicdt^d  operation,  being  just  the  reverse 
of  the  former.  Thus,  x  —  y^x"'.  That  is,  raising  x  to  the  with 
power,  and  then  indicating  the  wth  root,  leaves  the  value  represented 
unchanged. 

EXAMPLES. 
Ex.  1.  Reduce  '^dh^  to  a  form  of  a  radical  of  the  3rd 
degree. 

Operation.     7a  V  =  ^{la'x^y  =  ^343^. 

Explanation. — If  1  cube  7aV  and  then  extract  the  cube  root  of 
this  cube,  the  result  will  evidently  be  the  same  as  at  first.  Now 
ila^x^y  =  343aV.  But  instead  of  performing  the  operation  of 
extracting  the  cube  root  of  343aV,  which  would  evidently  return 
it  to  Ta^a;^,  I  simply  indicate  the  operation,  and  have  \^MZa^x*. 

2.  Reduce  2ay—S  to  the  form  of  a  radical  of  the  second 
degree.  Result,  V^aV— 12fl«/  +  9. 

3.  Reduce  a—x  to  the  form  of  the  cube  root. 


^/\ 


4.  Reduce  A  /  5  ^^  the  form  of  the  4th  root. 


3 
Reduce  -  to  the  form  of  the  3rd  root. 

0 


Result, 


184  CALCULUS   OF   RADICALS. 


6.   Eteduce  ^  to  the  form  of  the  4th  root. 


221.  Cor. — To  introduce  the  coefficient  of  a  radical  under 
the  radical  sign,  it  is  necessary  to  raise  it  to  a  power  of  the 
same  degree  as  the  radical ;  for  the  coefficient  being  reduced  to 
the  same  form  as  the^  radical  by  the  last  rule,  we  have  the  pro- 
duct of  two  like  roots,  which  is  equal  to  the  root  of  the  pro- 
duct  (194,  and  205). 

EXAMPLES. 

1.  Introduce  the  coefficient  in  SxV^x^  under  the  radical 
sign. 

Model  Solution. 
Operation.  dx\/2x^  =  ^/^fx"  x  \/2^=  ^^Ix^  x  ^x"  =  v^54^^ 

Explanation. — Cubing  ^x  I  have  27a5^,  the  indicated  cube  root  of 
which  is  'v/27x\  This  is  evidently  the  same  in  value  as  3a;.  Hence 
Zx^^x^  =  /v^27^^  X  \/2^  =  \/27^>^^  =  ^54^  (209). 

2.  Introduce  the  coefficient  in  -a/2  under  the  radical  sign  ; 
in  g-v/3;  in  ^Vi;  in  gv'O. 


Results,  a/^,  \/|,  \/j.  ^/l■ 


3.  Introduce  the  coefficients  in  the  following  expressions 
under  the  radical  signs  ;    So?  V2ax,  (a—x)  Va-i-x,  ^  V^a, 

. c\^ 

(x—y)  Vx—y,  and  ^v27a^. 


Ttvo  of  the  results  are  V{(i^—o^^)  (a—x),  and  V{x—yY. 


liKDUcTlOXS.  185 

4.  Introduce  under  the  radical  signs  the  coefficient  in  the 

following:  i^,  ^^,  ox^/2b^,  and  ^^^. 

The  last  two  are  v'5"»+*^-*,  and  \/  — -r- 

222,  Prob.  4.--To  peduco  radicals  of  different  degrees 
to  equivalent  ones  having  a  common  index. 

Rule. — Represent  the  numhers  hy  means  of  frac- 
tional indices.  B educe  the  indices  to  forms  having  a 
common  denominator.  Perform  upon  the  numbers 
the  operations  represented  hij  the  numerators,  and 
indicate  the  operation  signified  hy  the  detioniinator. 

Demonstration.— The  only  point  in  this  rule  needing  further 
demonstration  is,  that  multiplying  numerator  and  denominator  of 
a  fractional  index  by  the  same  number  does  not  change  the  value 

a  ma  a  111 

of  the  expression,  /.  t'.,  that  jf  =  jP'.     Now,  a^  =  .r^  x  a?  x  ic^ 

to  a  equal  factoi*8.     If  now  we  resolve  each  of  these  factors  into  m 
equal  factors,  x  will  be  resolved  into  nib  equal  factors,  and  one  of 

I  a 

them  will  be  represented  xP>.     But  as  in  .r^  there  are  a  factors,  each 
x^,  and  as  each  of  these  is  resolved  into  m  factors,  there  will  be  in 


all  ma  factors,  each  .r^.     Hence  3^  —  .r^. 

[When  f  is  used  as  a  common  traction,  wc  show  that  f  =  J 
thus:  I  signifies  3  of  the  4  equal  partx  of  some  quantity.  If  now 
we  separate  each  of  these  3  parts  into  2,  the  entire  quantity  will  be 
separated  into  8  parts,  and  in  the  three  parts  there  will  be  6.  /.  f 
=  f.  In  an  analogous  manner  when  |  is  used  as  an  exponent,  we 
show  that  f  is  equivalent  to  J,  thus :  f  as  an  exponent  signifies  3 
of  the  4  equal /<ictor«  of  some  quantity.  Now  if  we  resolve  each  of 
these  3  eqvL2\  factors  into  2,  the  entire  quantity  will  be  resolved  into 
H factors^  and  in  the  ^factors  there  will  be  0.  .*.  |  as  an  exponent 
is  equivalent  to  |.] 

EXAMPLES. 

Ex.  1.  Reduce  ^%a^  and  ^/^m^y  to  forms  having  a  com- 
mon index. 


186  CALCULUS   OP   RADICALS. 


Model  Solution. 

Operation.  -v/So^  =  (2a^a;)i  and  \/im^  =  (4w'y)i.  But 
{2a'x)^  =  (2a'xyl  =  (8a'x')^  =  ^8^V ;   and  (4w»i  =  (4wiV)t 

Explanation. — Representing  the  given  numbers  by  means  of  frac- 
tional indices,  I  have  (^a^x)^  and  {^rn?ij)^.  These  indices  reduced  to 
forms  having  a  common  denominator  are  f  and  |.  Now  (Sa'a;)^  signi- 
fies one  of  the  two  equal  factors  of  'ia^x ;  while  (3a^£c)t  signifies  ^A^-e^  of 
the  six  equal  factors  of  2a^a7.  Hence  (2«°a!)^  =  {^a^x)^.  In  like 
manner  (4m^y)^  signifies  <me  of  the  ^Ar^e  equal  factors  of  4mV;  while 
{^m^yY  signifies  two  of  the  six  equal  factors  of  the  same.  Hence 
{hn^yY  =  {irn^yY.  Finally,  as  (2a*«)t  is  the  same  as  the  6th  root 
of  the  cube  of  2a^cc,  I  have  ^/Wx^.  In  like  manner  (im^yY,  mean- 
ing the  6th  root  of  the  square  of  Avi'^y,  becomes  '\/'iQ7n*y'-. 


2.  Reduce  V2  and  ^3  to  forms  having  a  common  index. 

Results,  \/8  and  \/9. 

3.  Reduce  V3,   \^6,   \^2  to  forms  having  a  common 
index. 

4.  Reduce  \^2a^,  ^/^x,  and  \/xy  to  forms  having  a  com- 
mon index.  One  result  is  V^^da^. 

5.  Reduce  V^  and  Vhx^  to  forms  having  a  common 
index.  Results,  (a^a^Y  and  {l^x^)^. 

6.  Eeduce  a^,  {bh)^  and  (3c)^  to  forms  having  a  common 
mdex.  Results,   ^V^T^,  V^^  ''^625"j^ 

7.  Reduce  a?»  and  i/^  to  forms  having  a  common  index. 

8.  Reduce  2c V^  and  6a\^2y  to  forms  having  a  common 
index. 


ItKDtJCTlOXS.  IBY 

Suggestion. — The  radical  factors  can  be  reduced  to  forms  hav- 
ing a  common  index  without  affecting  the  coefficient,  since  the 
operation  does  not  affect  the  value  of  the  radical. 

9.  Reduce  4'V^5a;2;y,  2\/2xy,  and  lOaVSbx  to  forms  having 
a  common  index.     Results^  2\^2xy,  10a\^2Wx^,  4v^2Scy. 

10.  Reduce  a-\-c  and  {a—c)^  to  forms  having  a  common 
index.  Results,  (a^-\-2ac  +  (^)^  and  (a—c)K 


^i23.  Prob.  5. — To  reduce  a  fraction  having  a  mono- 
mial radical  denominator,  or  a  monomial  radical  factor  in  its 
denominator,  to  a  form  having  a  rational  denominator. 

Rule.  —  Represent  the  radical  with  a  fractional 
index,  and  then  multiply  both  terms  of  the  fraction 
hy  the  quantity  in  the  denominator  with  an  index 
which  added  to  the  given  index  makes  it  integral. 

Demonstration. — Since  two  factors  consisting  of  the  same  quan- 
tity affected  by  the  same  or  different  exponents  are  multiplied  by 
adding  the  exponents  (90),  and  the  sum  of  the  exponent  of  the 
denominator  and  the  factor  by  which  we  multiply  it  is  an  integer, 
the  product  becomes  rational.  The  value  of  the  fraction  is  not 
altered,  since  both  its  terms  are  multiplied  by  the  same  number. 

EXAMPLES. 

3(1  ^2x 
Ex.  1.  Reduce  — nz^  to  a  form  having  a  rational  denom- 

inator. 

Model  Solution. 

Operation.      ?«^  ^  3a(2^x  (3x)i^  3<6^i  ^  ^^^ 
^/Zz  (3«)ix(3x)i  8.r 

Explanation. — Using  fractional  exponents,  I  have — •     Since 

multiplying  numerator  and  denominator  by  the  same  number  does 
not  alter  the  value  of  the  fraction,  an  1  as  (Sx)^  x  (3x)i  makes  3a*,  I 


188  CALCtTLirg    OF     RADICALS. 


can  rationalize  the  denominator  of  this  fraction  by  multiplying  both 

its  terms  by  (3aj)i.     This  gives  — ^ — ~  which  reduces  to  a\/Q.   [If 

ox 

the  rationale  of  these  last  reductions  is  not  perfectly  familiar  it  should 
be  given.     Thus  (6.T'^)i=(6)i(aj'')i=( 
and  cancelling  the  dx  I  have  «\/6.] 


be  given.     Thus  (6..^)i=(6)V)i=6ia^,  whence  ?^^^  =  ^^, 


2 

2.  Rationalize  the  denominator  of 


2  2 

Suggestion.     ~~p=  =  —-^ ,  and  ai  is  the  factor  which  rational- 
3  V  «'      3a^ 
izes  it. 


Uesult,  -x-ic' 


3.  Rationalize  the  denominator  of  — —  • 


4.  Rationalize  the  denominator  of  -g-^-     Result,  z^.—  , 

Va  a 

5.  Reduce  -rpto  a  form  having  a  rational  denominator. 

ya 

Result, 

6.  Reduce    — =:,  -r-=,  -ir7=>  and  ——    to    forms    havinsr 

rational  denominators.  Owe  of  the  results  is  _/y/i6. 

Z 

\/%       (3)ix(6)i       Vis       SV^       1    /5 
Suggestion.     -^  =  —  ^         ,  =  ^  =  -^  =  sV^. 
V^       (6)ix(6)i  6  6  2^ 

Scholium. — This  process  is  equally  applicable  to  any  form  of 
radical  factor  in  the  denominator,  whether  monomial  or  polynomial. 


7.  Rationalize  the  denominator  of  '- 

wa  +  X 

QMnnn.#!««         V^"^  _,   {(l-X)\  ^  {a-^X)\  _   ^J o" -X^ 

Suggestion.        ,         = \ 7  =     ^  ,  ^    - 

\/a^-x       0?  +  a;)ix(«  +  5cp  ^"^* 


REDUCTIONS.  189 

8.  Reduce  — r; .-  to  a  form  having  a  rational  denom- 

inator.  Result.  — z~ — ->  .,      » 

6c— 2x^ 


224.  Prob.  6. —To  rationalize  the  denominator  of  a 
fraction  when  it  consists  of  a  binomial,  one  or  both  of  whose 
terms  are  radicals  of  the  second  degree. 

Rule. — Multiply  both  terms  of  the  fr  actio  ft  by  the 
denominator  with  one  of  its  signs  changed. 

Demonstration. — This  rationalizes  the  denominator,  since  in  any 
case  it  gives  the  product  of  the  sum  and  diflFerence  of  the  two 
terms  of  the  denominator,  which  being  equal  to  the  difference  of 
their  squares,  frees  either  or  both  from  radicals,  as  the  square  of  a 
square  root  is  rational. 


EXAMPLES 

of 


Ex.  1.  Rationalize  the  denominator  of — ^ 


Operation. 


Model  Solution. 

a{a  +  \^h)        _  a'  -I-  a  ^/b 


a—  ^ft       (a-  y^)(a  +  y  ft)  «'  -^ 

Explanation. — I  observe  that  a—'s/hvnW  be  rationalized  by  mul- 
tiplying it  by  o+  \/^  since  the  product  of  the  sum  and  difference 
of  two  quantities  is  the  difference  of  their  squares.  Hence  multi- 
plying both  terms  of  the  fraction,  so  as  not  to  alter  its  value,  I  have 

tt'-f-av/ft 

2.  Rationalize  the  denominator  of     _      — - 

va—Vh 

a—b 


190  CALCULUS   OF   RADICAL^?. 

3.  Reduce —to  a  form  having  a  rational  denom- 

2a/2— aVa 

inator. 

,,  3(2\/2  +  3\/3)  _  6a/2  +  9\/3  6a/2  +  9a/3 

^^^^^^'  8^2T~    ~        -19        ^^  19 

4.  Reduce —  to  a  form  havius  a  rational  denorai- 

3-2V2 

nator.  Result ^  4+^/3. 

1-_V5    I  +  V2 


5.  Rationalize    the    denominators  of 


— ;=,  and 7=^' 

2V6— Vl8  Va  +  ^+V«— aJ 


3  +  ^5'  2  +  V2' 


i?^5i^/jfs,  2 -a/5,  iV2^  9  +  l-A/iO,  and  ^     ^^^  — ' 


6,  Rationalize  the  denominator  of 


Result^ 


x-^1 


FOR  REVIEW  OR  ADVANCED  COURSE. 

225,  Prob.  7.— A  factor  may    be    found    which    will 
rationalize  any  binomial  radical. 


Demonstration. — If  the  binomial  radical  is  of  the  form  \/{a  ■^h)'!^^, 

m  n — m 

or  (a +5)*,  the  factor  is  (a  +  h)  «  ,  according  to  (223). 

If  the  binomial  is  of  the  form  ■y/as-\-  yJ'  or  «"»  +  &«,  let  a'"  —  x, 


REDUCTIONS.  191 


and  ft»  =  y ;  whence  w^  =  x^,  and  ft*  =  y\  Also  let  y  be  the  least 
common  multiple  of  m  and  n,  whence  -xf  and  y'^  are  rational.     But 

ifr  —  a'",  and  y^  =  J*.  If  now  we  can  find  a  factor  which  will  ren- 
der  a^  +  y,  a?'^±y^  this  will  be  a  factor  which  will  render  am  +  ?»", 
rr'"±7>'»  which  is  rational.  To  find  the  factor  which  multiplied  by 
,*^y'  gives  a^P  ±  y'/*,  we  have  only  to  divide  the  latter  by  the  former. 

Now ^=  a-»(i>-l>— a^(J>-2)y' +aj»(l>-3)y2)-_a-»(p-4)y3r^ ±y'(p-l) 

(4),  the  +  sign  of  the  last  term  to  be  taken  when  y  is  odd,  and  the 
—  sign  when  it  is  even  (126).  Therefore  a!»^p-''— af<P-2)y +  af<P-«>y2r 
_2^p— i)y8r^ ±y'^<i>-i>,  is  a  factor  which  will  render  ^a'-\-  ^Jf 

«  r 

rational,  of  being  understood  to  be  a^^  and  y  =  6»,  and  p  the 
L.  C.  M.  of  m  and  n. 


If  the  binomial  is  J^a*  —  ^^b%  the  factor  is  found  in  a  similar 
manner,  and  is  aj»(i'-i)+«^^2^y''+a5»(^3)y2r+ +y(i>-i). 


EXAMPLES. 
Ex.  1.  Rationalize  Va-\- \/b  or  a^-^h^. 

Solution.— In  this  example  «=1,  /•=!,  i>=6,  a;=a»,  and  y—h  ■ 
Hence  formula  (A)  becomes  ah—a^lik  +  a^ht—ab  +  azh^—hs. 

This  factor  multiplied  into  a^  +  ft^  gives  a"— &',  as  the  rational- 
ized product. 

2.  Find  a  factor  which  will  rationalize   V^  —  \/t^  or 

Result,  eV-4.A^f  +  6^t;t-feV-t;l4.eVt;¥+eVt;V  +  ^Vt;¥ 
_j.eft;V^  +  efy¥  +  elv¥-|-efv^-fv^,  is  the  required  factor, 
and  the  expression  rationalized  is  e®— v®. 


192  CALCULUS   OF   RADICALS. 

226.  Prob.  8. — A  Trinomial  of  the  form  Va  -f  ^/h  -\-  Vc 
7nay  he  transformed  into  an  expression  with  hut  one  radical 
term  by  multiplying  it  by  itself  with  one  of  the  signs  changed, 
as  Va+A/b  —  Vc.  ITie  product  thus  arising  may  then  he 
treated  as  a  hinomial  radical  hy  conddering  the  sum  of  the 
rational  terms  as  one  term,  and  the  radical  term  as  the  other. 

Thus,  {Va-{-Vb-^Vc){Va+Vb—Vc)=a-^b—c  +  2Vab. 

Again,   [{a-\-b—c)-j-2\/ab]x[(a-\-b—c)—2Vab]  =  a^  +  U^ 
-\-(^—2ab  —  'Zbc~2ac,  a  rational  result. 

Ex.  1.  Rationalize  VS—Vl  —  V^-  Result,  4. 


COMBINATIONS    OF    RADICALS. 


ADDITION   AND    SUBTRACTION. 

227.  Prob.  1. — To  add  or  subtract  radicals. 

Rule.  — //  the  radicals  are  similar  the  rules  already 
given  (72,  77)  are  siijficient.  If  they  are  not  similar 
make  them  so  hy  (217-233),  and  eomhine  as  hefore. 
If  they  cannot  he  made  similar,  the  comhinations  can 
only  he  ind^icated  hy  connecting  with  the  proper  signs. 

Demonstration. — When  the  radicals  are  similar  the  radical  factor 
is  a  common  quantity  and  the  coefficients  show  how  many  times  it 
is  taken.  Hence  the  sum,  or  difference,  of  the  coefficients,  as  the 
case  may  be,  indicates  how  many  times  the  common  quantity  is  to 
be  taken  to  produce  the  required  result. 

If  the  radicals  are  not  similar,  the  reductions  do  not  alter  their 
values;  hence  the  sum  or  difference  of  the  reduced  radicals,  when 
they  can  be  made  similar,  is  the  sum  or  difference  of  the  radicals. 


COMBINATION — ADiHTiON    AND  SUBTRACTION.        193 
EXAMPLES. 

Ex.  1.  Add  Vl^  and  \/242. 

Model  Solution. 
Operation.     ^18  =  2^/2,  and  ^^42  =  11^2. 

.-.  V18+ V242  =  ^^/2  +  ny^2  =  14^2. 


Explanation.  \/l^  =  V^  ^  2.  But  the  square  root  of  the  pro- 
duct equals  the  product  of  the  square  roots ;  hence  V^  ^  2=3/y/2. 
In  liiie  manner  v'^242  =  ^^121  x  2  =  11^2.  Therefore  ^18 
+ 'Y/242  =  3^^2+11^/2.  But  three  times  any  quantity,  as  \/2, 
and  11  times  the  same  are  14  times  that  quantity.  .'.  3Y^2  +  liy'2 
=  14^2. 


2.  Add  V2Wx^  and  VT92xf.  Sum,  i7yVSx, 

3.  Add  \/500  and  v^lOS. 

4.  Add  Va^y  and  Vc^y-  Sum,  (a  +  c)Vy» 

5.  From  \/605  take  the  V^OS.  Diff.,  2\/5. 

6.  Add  V605  and  —  \/405. 

7.  Add  3^/^  and  2y^l. 

8.  From  3a/^  take  -3y  ^-  />ijf.,  ^\/lO. 

9.  What  is  the  sum  of  \/  j  and  \ /q-''      ^^.,  ^V3' 


10.   What   is   the   difference  of    V2a3^  —  iax  +  2a  and 
V2aa?^H-4aa;-t-2a?  viw-s  2V2a. 


194  CALCULUS    OF    RADICALS. 

Why  should  no  sign  be  given  to  the  last  answer  ?  If  the  problem 
read,  From  -\/2cta;^— 4a«+2a  take  \/2a^T^ax  +  2aj  why  would  the 
answer  be  —2^ 2a  ? 

11.  What  is  the  sum  of  (a—xy  Vxy  and  {a-{-xY\/xy? 

Ans.,  2(fl2+ic2)V^. 

12.  What  is  the  difference  between  (a—xf^/xy  and 
(a-^xYVxy"^  Ans.,  Aax^yi 


3 

5' 

Sum,  WS. 


13.  Find  the  sum  of  8\/|,  V60,  _2iVl5,  and  a/ 

14.  From  W~  take  W 

.,  (3«.-|)v/; 


W 
Mh 


Rem.,  [Zax—~\\y-^- 


16.  Add  «Y^l+[^]^  and  ^\/l  +  [ff  • 
Suggestions.    a/^J  =  a>^l+|  =  ay^^ 

«f  Vat  +  6i     In  like  manner  &|/l  +[^1^  =  &t  Vof  +  fti 
The  sum  is  (at +  5f)Val+5f  =  (at+&t){al+&t)i  =  («l  +  &t)i 


16.  From  (a-o;)  V^F^^  take  (a-a:)  a /^±^. 

Rem.,  (a—x—1)  Va^—x^, 

17.  Add and  ; ^wm,  2a;. 

ay+Va:^— 1  a:— Va;^— 1 

18.  Add^^^  +  ^^^  and  ^^S-^^S- 


COMBINATIONS— Mr  LTIPLICATION.  196 

MULTIPLICATION. 

228,  Prop.  1. — The  product  of  the  same  roots  of  two  or 
more  quantities^  equals  the  like  root  of  their  product. 

Demonstration. — That  is  \/x  x  \^y  =  ^xy.  This  is  evident 
from  the  fact  that  \/xy  signifies  that  xy  is  to  be  resolved  into  m 
equal  factors.  If  now  x  and  y  be  separately  resolved  into  m 
equal  factors  and  then  one  factor  from  each  be  taken  to  make  a 
group,  there  will  be  m  such  equal  groups  in  xy.  Thus  ^x  is  one 
of  the  m  equal  factors  of  x,  and  y'y  is  one  of  the  m  equal  factors 
of  y.  Hence  \\/x  x  y'y]  •  \/\/x  x  -y^y]  •  [/{/«  x  ^y'\  etc.,  to  m  fac- 
tors of  ^x  X  ^y,  makes  up  xy.  Therefore  ^x  x  \/y  =  ^vy. 
(See  Abts.  205  and  202.) 


22^,  Prop.  2. — Similar  Eadicds  are  multiplied  by  multi- 
plying the  quantities  under  the  radical  sign  and  writing  the 
product  under  the  common  sign  ; 

Or  by  indicating  the  root  by  fractional  indices,  and,  for  the 
product,  taking  the  common  number  with  an  index  equal  to  the 
sum  of  the  indices  of  the  factors. 

Demonstration.  1st.— Since  similar  radicals  are  the  same  root 
of  the  same  quantities,  as  \^x  x  y'a;,  this  is  only  a  pailicular  case 
under  Prop.  1. 

2nd,  «"•  X  a?"  signifies  that  one  of  the  m  equal  factors  of  x  is  to  be 
multiplied  by  another  of  the  m  equal  factors,  or  by  itself.  This 
gives  2  of  the  m  equal  factors  of  x,  which  is  what  is  indicated 
by  a?». 


230.  Prob.  2.— To  multiply  radicals. 
Rule. — //  the  factors  have  not  the  same  index,  re- 
duce them  to  a  common  index,  and  then  multiply 


196  CALCULUS   OF   RADICALS. 

the  mvmbers  under'  the  radical  sign  and  write  the 
product  under  the  common  sign. 

Demonstration. — (This  is  the  same  as  Prop.  1.) 

EXAMPLES. 
Ex.  1.  What  is  the  product  of  V2  and  ^/^  ? 

Model  Solution. 
Operation.     \/2  =  \/8,  and  v^3  =  ^9.    .-.  '\/2  x  \^S  =  ^ 
x^9  =  ^72. 

Explanation.  '\/2  =  ^y^S,  since  the  former  is  one  of  the  two 
equal  factors  of  2,  and  the  latter  is  three  of  the  six  equal  factors 
of  3.  In  like  manner  \^d  ~  v^9.  Consequently,  ^/2x  ^y/S  —  ^/S 
X  -y/O.  Now  since  the  product  of  the  same  root  of  two  numbers  is 
equal  to  the  like  root  of  the  product,  /y/8  x  ^^  =  ^^72. 


2.  Multiply  \/dac  by  \/2ac.  Prod.,  V433«V. 

3.  Multiply  ^/a—x  by  "s/ a—x. 

Prod.,  ^a^—hahc ^AS)a^x^—  VWx^  +  hax^—7^. 

4.  Multiply  y  ^  by  y  ^.  Prod.,  \ 

6.  Multiply  ^/l  by  \/l. 

Prod.,  3^+^  =  3^  =  ''v/6561. 

6.  Multiply  "V^Zax  by  '\/2ax. 


Prod.,  (2aa;)H  =  ""^Z A^Q^Qa^x^. 
7.  Multiply  ^Jl  by  ^^\ 


Prod,   A/|=:i^486. 


5.  Multiply  3a/2S  by  2^xy, 


COMBINATIONS — M  ULTlPLlCATIOK.  W 

Suggestion. — Here  we  have  the  continuous  product  of  3,  \/2aT^ 
2,  and  yiri/.  But,  as  the  order  of  the  factors  is  immaterial,  we  may 
write  3  X  2  X  V^o*  x  ^xy=Q  ^/Sa^xY- 

9.  Multiply  3a^|  by  ^y\'  Prod.,  6'\/236196. 

10.  Multiply  6a^  by  3ai 

11.  Multiply  2\/ab  by  d\^ab.. 

12.  Multiply  ia^b^  by  ba^bl 

13.  Multiply  Sx^y^  by  2x^y^  and  represent  the  product 
without  fractional  exponents. 

14.  Multiply  A /-  by  \V^  and  represent  the  product 
without  the  use  of  the  radical  sign  and  in  its  simplest  form. 


Prod.,  ^(9000)1 


15.  Multiply  cr  by  b". 


Prod.,  a^b*  or,   ^/aP'b'^,  or  («^^)'»». 

16.  Multiply  |V5    by  ^\^iO.  Prod.,  ^v^250. 

17.  Multiply  av^a;,  b\/y,  and  c^z  together. 


Prod.,  ahr^xf'Py^Pz'^^ 

18.  Show  that  2\/9  x  16  =  16^/13. 

19.  Show  that  ^^24  x  6v^3  =  6\/l2. 

20.  Multiply  2\/a— 3v^  by  SVa  +  Wb. 

Operation. 

2  \/a  —  3  yT 

3  \/a  +  2  v^ 

I  6a        —  9  \/c^ 

+  4  \/qft  —  6ft 
6a       —  5  \/ab  —  6&,  or  6(a— J)— S'y/aJ. 


198  CALCt^LltS   OP   RADICALS. 

21.  Multiply  3  + a/5  by  2-\/5.  Prod.,  l-^V?. 

22.  Multiply  \/2-fl  by  V^-1.  Prod,,  1. 

23.  Multiply  iW^-Wl^  by  Vs  +  Vs.   . 

Pro^.,  2^^3  —  ^/10. 

24.  Multiply  V^12  +  Vl9  by  '^12-a/T9.       Prod.,  5. 
By  (22§)  ^12+ Vl9x  ^13-^19  =  ^ {\^+ \/\^){l^-^/\^). 

25.  Multiply  ^2_^V2  +  1  by  a^-\-a\/^  +  l. 

Prod.,  aHl- 

26.  Expand  (x^  +  l){x^—xV^^\)(x^^xV'^-\-l). 

Prod.,  cc6+l. 

27.  Multiply  3a/45-7V5  by  Vii+2\/9i.    Pro6?.,  34. 

28.  Multiply  ^/a-\-c^/l  by  Va~c\/h. 

Prod.,  a-^c^^. 


DIVISION. 

;^5i.  Prop. — The  quotient  of  the  same  roots  of  two  quanti- 
ties equals  the  like  root  of  their  quotient. 

Demonstration. — Let  m  be  any  integer  and  x  and  y  any  numbers ; 
we  are  to  prove  that  v^^-^v^y,  or  5^  =  4/^.     Now,  that  ^' 

=  y  -  is  evident,  since  ^^  raised  to  the  Twth  power,  that  is 

y  ^y 

Ux X  yx  X  ya;  X  ytc to  w  factors      x 

-iTF- — ^- — ^;:7- — ^i7= ==-;     whence    it    appears 

y  2/  X  yy  x  y  2/  x  y  y to  m  factors      y 

that  ^^  is  the  rnth  root  of-  or  equals  i/  -.     (Akts.  206,  202.) 

\/y  y  ^  y 


COMBINATIONS— DIVISION.  199 

232.  Prob.  3.— To  divide  Radicals. 

Rule. — //  the  radicals  are  of  the  same  degree, 
divide  the  number  under  the  sign  in  the  dividend 
by  that  under  the  sign  in  the  divisor,  and  affect  the 
quotient  witlo  the  common  radical  sign. 

If  the  radicals  are  of  different  degrees,  reduce 

them  to  the  same  degree  before  dividing. 

Demonstration. — [Same  as  above ;  or,  it  may  be  considered  as 
the  converse  of  the  corresponding  case  in  multiplication.] 

EXAMPLES. 

Ex.  1.  Divide  V^cfif  by  ^iay. 

Model  Solution. 
Operation.     a/3«V  =  \/27^y,  and  v^2^y  =  >v/4^'. 

Explanation.— Since  V^^V  =  V^27aY,  and  ^2ny  =  ^4aY 
.  _    =   l-^-—-'    And  since  the  quotient  of  the  6th  root  of 

two  numbers  is  the  6th  root  of  their  quotient,  -^ ^   =  V  zlJl-^ , 

which,  by  performing  the  operation  indicated,  becomes  y  — -—  ; 
and  by  reducing  so  that  the  number  under  the  radical  sign  shall 
have  the  integral  form,  this  becomes  ^  ^'kd2a*i/\ 

2.  Divide  VV25ah^y  by  ^hahcy.  Qvot.,  5\/ax. 

3.  Divide  a/3  by  ^.  Quot.,  ^^^  =  ^\/^^. 

4.  Divide  ^J\  by  >y/|.  Qmt.,  \^mA. 

5.  Divide  V^^^  by  \/2a^.  QuoL,   '\^I^. 

6.  Divide  ^/72  by  a/2.  Quot.,  \/l. 


^00  Calculus  of  radicals. 

7.  Divide  x^y^  by  x^y^,  and  represent  the  quotient  with- 
out fractional  or  negative  exponents.  Quot.,  \  / -. 

V  y 

8.  Divide  24:Vay  by  6v^ay. 

Suggestion.    '-^^^  =  4^  ^  4-^5  =  4^^^^  =  4i/I 

9.  Divide  126V^y^  by  26\/^^,  representing  the  quo- 
tient without  the  radical  sign.  Quot,  5x^y~K 

10.  Divide  a/6  by  Vi.  Quot,  V3V2. 

11.  Divide  20v^200  by  4^2.  Quot,  b'^l. 

12.  Divide  ^^  by  ^f. 

13.  Divide  iVi  by  V^  +  SVf 

Suggestions.     ^/2  +  8  \/^  =  2  ^/\  +  3  Vi  =  5  Vf.    Wbence 

VS  +  SVi       5V^       5       10 

14.  Divide  v^a^H^  by  v^^II*.  C^^ojJ.,   v^^+^. 

15.  Divide  {aWcf^  by  («J)i  G«^Oif.,  a^/lc. 

16.  Divide  200  by  \/40.  0«^oJ^v  lOVlO^  or  (10)1 

17.  Divide  aVx—\/hx-\-aVy—Vhy  by  V^+Vy. 
Suggestion. — Observe   that  a  \fi—  '\/hx  +  a  \/y—  \/by  =  a  {y/x 

+  \/y)-VKV^+Vy)  =  {a-V^{V^+Vy)' 

18.  Divide  a  +  h-^c  +  2Vab  hj  Va-\-V^—Vc, 


COMBINATIONS— INVOLUTION.  201 

Operation. 

a  +  'it^/ab  +  b  —  c  \  \/a  +  \/b  —  \/c 
a  +     \/ab  —  \/ac      ^a  +  \/ft  +  yc 


^ab  +  b  —  \/bc 


\/ac  +  \/bc  —  e 


19.  Divide  bV^^^  by  aVia-^-by. 

20.  Divide  ^/c^—a?  by  rt— ar.  ^wo/.,  v/-^-:- 

21.  Dmde  my/— ^  by  ^  —  •      G^.o^.,  -J^. 


INVOLUTION. 
233.  Prob.  4.— To  raise  a  radical  to  any  power. 

Rule. — Involve  the  coefficient  to  the  required  power, 
and  also  the  quantity  under  the  radical  sign,  writing 
the  latter  under  the  given  sign. 

Demonstration.— This  results  directly  from  the  principles  of 
multiplication  of  radicals  (230).  Thus,  to  raise  a/^b  to  the  7»th 
power,  is  to  take  m  factors  of  ay'b^  which  gives  « y  6  x  a-y^J  x  ay^ft, 
L'tc,  to  m  factors.     But  as  the  order  of  the  factors  is  immaterial 

(§5)  this  may  be  written  aaa to  wi  factors  x  -{/&  x  ^yb  x  \/b  to 

m   factors.      But  ana to   m  factors  is   by   notation   a"*,   and 

V^6x  v^x  v^6  -  -  -  to  m  factors  is  by  (230)  v^.     .*.  The  wth 
power  of  a\^&  is  ^"•'y^.    q.  e.  d. 

EXAMPLES. 
Ex.  1.  Raise  Ja/|  to  the  3rd  power. 


202  CALCULUS   OF   RADICALS. 

Model  Solution. 
Operation.  J^VW  =  iVf  •  Wh  Wi  =  i  •  i  "  i  x  VI  Vl 

Vl  -  sV  VtIt  or  Tfy Vl  =  -rfirVlO. 

Explanation. — The  cube  of  ^■\/l-  is  a  product  three  factors  each 
=  aVIi  ^u^  as  the  order  of  the  factors  is  immaterial  this  may  be 
considered  as  3  factors  each  =  ^,  or  -^j,  and  3  factors  each  =  yf , 
or  f  Vf  •  Hence  a  Vf)«  :=  ^  x  |  ^f  =  ih  VI  Which  by 
removing  the  denominator  from  under  the  radical  sign  becomes 

2.  Raise  2a/3^  to  the  second  power. 

1  3 

3.  Raise  -y4:X^y  to  the  6th  power. 

4.  Raise  —  3a/^  to  the  3rd  power. 

5.  Square  3  —  ^2.  /S^ware,  11— 6a/2. 

6.  Cube  a/2- a/3.  *  Cube,  ll\/2— 9\/3. 

7.  Cube  2Vx—y. 

Cube,  S{x—y)Vx—y,  or  8(2^—1/)^. 

^54.  Cor. — To  raise  a  radical  to  a  power  ivJiose  index  is 
the  index  of  the  root,   is  simply  to  drop  the  radical  sig7i. 

Thus,  the  square  of  Vab  is  ab,  the  cube  of  V^x^y  is  "^x^y, 

the  square  of  V— 2a^^  is  —^a^b,  the  5th  power  of  "s/a^—l^ 
is  d^—b^. 

Scholium. — This  process  of  involution  is  a  special  case  under 

EVOLUTION. 

23r^»  Prob.  5. — To  extract  any  required  root  of  a  mono- 
mial radical. 

Rule. — Extract  the  root  of  the  coeffident,  and  of  the 
quantity  under  the  radical  sign  separately,  affecting 


C0MBINATI0K8 — EVOLtJTiOX.  20t'i 

the  latter  with  the  giveii  radical  sign.     Reduce  the 
result  to  its  simplest  form. 

Demonstration — The  nt\\  root  of  a-y^j,  signifies  one  of  the  n 
equal  factors  oi a^^h.  Hence  if  we  resolve  a  into  n  equal  factors, 
and  y  6,  into  n  equal  factors,  a  group  consisting  of  one  from  each 
is  tae  wth  root  o{  a^h.  Thus  one  of  the  n  equal  factors  of  a  is 
V^a,  and  of  y  ^  V^y  ^  **or  \^>^lx  V^^ft  x  \^^b-  --ton  factors, 
is  \^^h  b'l-bton  factors  =  ^&  (233,  231).  If  now  we  take  a 
group  consisting  of  one  from  each  set  of  factors,  that  is  y^  V  y  b, 
we  have  the  7ith  root  of  a  y'ft,  since  we  have  one  of  its  7i  equal 
factors.     Q.  B.  D. 

EXAMPLES. 
Ex.  1.  What  is  the  3rd  root  of  WSc^? 

Model  Solution. 
Operation.    V^4  \/daFx  =  \/4'i^^/^a^x  =  y^4y^3a^ 

Explanation. — The  cube  root  of  4:\/Sa^x  is  one  of  the  3  equav 
factors  which  compose  it.  In  order  to  find  this,  I  resolve  each  of 
the  two  factors  4  and  \/da^x  into  3  equal  factors,  and  take  one  of 
each.  4  resolved  into  3  equal  factors  becomes  y^4  x  \^i  x  -^4. 
(A  process  which  in  reality  is  only  indicated.)  In  like  manner  \/3a*«, 
resolved  into  3  equal  factors  becomes  V  \/da^xx  \/da^xx  \/da*x 
or  V^Sa'xx  Vy'^^xx  V ^  Sa^x  since  the  root  of  the  product 
equals  the  product  of  the  roots.  Now  taking  one  factor  out  of  each 
of  these  groups  I  have  y^fv -y/aia'i,  and  as  three  such  factors  could 
be  formed  from  the  number,  \/4V'^'^*x  is  the  cube  root  of  4  \/Sa*x. 
But  this  expression  can  be  reduced  to  a  more  simple  fonn,  by  observ. 
ing  first  that  the  square  root  of  the  cube  root  is  the  6th  root,  and  then 
reducing  ^l  to  the  same  radical  form.  Thus  I  have  \^4  V ^da*x 
=  'V^^S^  =  ^16  ^da^  =  ^4S^  by  (233). 


204  CALCULUS  OP  KADICALS. 

2.  Extract  the  cube  root  of  a/^*^^  Root,  x^c^. 


3.  Extract  the  square  root  of  32v  192a8:c2. 
Suggestion.— The  square  root  of  32  v^IOSo^  =  4  ^/a  x  /v/l92aV 

4.  Extract  the  cube  root  of  -a^^/h. 

8 

5.  Extract  the  4th  root  of  l^c^^/x.  Root,  2\/a^. 

6.  Extract  the  cube  root  of  (a  +  x)\/a+x. 


Root,  va-\-x. 

7.  Extract  the  cube  root  of  :^\/  -o'  Root,  ^VSa. 

8.  Extract  the  square  root  of  ^\/^'  Root,  «v^2. 

236*  Scholium. — This  operation  is  but  a  special  case  of  affecting 
a  quantity  with  any  given  exponent  (194),  and  the  examples  may  be 
performed  according  to  the  rule  there  given.  Thus,  to  extract  the 
cube  root  of  Q^dax'^ ;  putting  it  in  the  form  8  x  (Sax^)i  and  divid- 
mg  exponents  by  3  (multiplying  by  ^)  we  have  8^{Saa!'^)l  =  2^Saa'. 
The  pupil  may  solve  the  following  in  this  manner : 

9.  Extract  the  5th  root  of  V32^. 

Suggestion.  -^/SSa;'"  =  (32)ia^.  Multiplying  exponents  by  ^  we 
have  (32)^'^a;.  But  (32)^^  =  \/(32)i  =  >^2.  .-.  \/ ^^/^2^  =  x^2. 
But  the  most  simple  way  to  solve  this  particular  case  is  V  y'SS*" 
=  V^32^  =  ^2x^  =-  x^y2.     Or  by  the  rule  given  (235). 

10.  Extract  the  square  root  of  \^4:9aK 

11.  Extract  the  cube  root  of  64^^80*. 


COMBINATIONS— EVOLUTION.  205 

12.  Extract   the   5th  root  of  486fl\^4rt2.   Root,  3v^. 


FOR  REVIEW  OR  ADVANCED  COURSE. 

237.  Prob.  6.— To  extract  the  square  poot  of  a  binomial, 
one  or  both  of  whose  terms  are  radicals  of  the  second 
degree. 

Solution. —Such  binomials  have  either  Jhe  fonn  a±;i\/5or 
m^^±  n^h.  Now  observing  that  {x  ±  yf  —  x^  ±  2xi/  +  y^,  we  see 
that  if  we  can  separate  the  first  term  of  any  such  binomial  surd  into 
two  parts  the  square  root  of  the  product  of  which  shall  be  ^  the 
other  term,  these  two  parts  may  be  made  the  first  and  third  terms 
of  a  trinomial  (corresponding  to  x'^-\-2xy  +  y^),  and  the  middle  term 
being  the  second  term  of  the  given  binomial,  the  square  root  will 
be  the  sum  or  difference  of  the  square  roots  of  the  parts  into  which 
the  first  term  is  separated. 

EXAMPLES. 

Ex.  1.  Extract  the  square  root  of  87— 12\/42. 

Solution. — Let  x  =  one  of  the  parts  into  which  87  is  to  be  sepa- 
rated, and  87— a;  the  other.  Then  we  have  \/x{S7—x)  =  — 6\/42, 
or  squaring,  Slx—x"-  =  1512,  or  «»— 87a;  =  —1512.     .*.    «=  63  or 


24.  and  we  have  V87-12V'42  =  V63-12\/42  +  24.  Now  \/Qd 
—  3\/7  and  '\/24  =  2\/6,  and  as  the  middle  term  of  the  trinomial 
is  negative  and  twice  the  product  of  these  roots,  its  square  root  is 
3  V7-2\/^-     (123.) 

2.  Extract  the  square  root  of  3\/6-f-2\/i2. 

Suggestion.  3^/6  is  to  be  separated  into  two  parts.  Let  them 
be  X  and  3\/6— «•  Then  ar(3\/6— ar)  =  12.  Whence  x  =  2\/6  or 
V^6,and  V3V6  +  2Vl2=V2V6  +  2yi2+y6=V2y6  +  V'V/Q 
=  </24+^6, 


206  CALCULUS   OF   RADICALS. 

3.  Extract  the  square  root  of  12 — Vl40. 

Root,  a/?— \/5. 

4.  Extract  the  square  root  of  11  +  6  \/2. 

5.  Extract  the  square  root  of  13  +  2a/30. 


6.  Extract  the  square  root  of  ax—2avax—a^. 

Suggestion. — Letting  y  be  one  of  the  parts  into  which  ax  is  to  be 
separated,  the  equation  from  which  its  value  is  found  is  y^  —  axy 
=  —a^{ax—a^)  or  — a^aj  +  cc*.     Whence  y  —  ax—d^  or  d\  and  the 


parts  are  ax— a?  and  a^.    Hence  yax—^^/ax—a^ 

a=  V  (ax— a'^)  — 2a\/"^— *^  +  *''  =  's/ox—fi^'^—^/a?  or  a— -x/oa;— a'. 


7.  Extract  the  square  root  of  2a  +  2va2_§2^ 


IMAGINARY  QUANTITIES. 

238 »  An  Imaginary  Quantity  is  an  indicated  even  root 
of  a  negative  quantity,  or  any  expression,  taken  as  a  whole, 
which  contains  such  a  form  either  as  a  factor  or  a  term. 

Thus  ^/^,  V^^S  W^^\  2+a/^  v^-e,  3-^^ 
etc.,  are  imaginary  quantities. 

230*  Scholium  I. — It  is  a  mistake  to  suppose  that  snch  expres- 
sions are  in  any  proper  sense  more  unreal  than  other  symbols.  The 
terra  Impossible  Quantities  should  not  be  applied  to  them  :  it  con- 
veys a  wrong  impression.  The  limits  of  this  work  prevent  anything 
more  than  a  mere  explanation  of  the  method  of  multiplying  or 
dividing  one  imagiaary  of  the  second  degree  by  another. 

240,  Scholium  2. — A  curious  property  of  these  symbols,  and 
one  which  for  some  time  puzzled  mathematicians,  appears  when 
we  attempt  to  multiply  '\/—a  by  y  — a.  Now  the  square  root  of 
any  quantity  multiplied   by   itself,  should,  by   definition,  be  th^ 


COMBINATIONS   OF    IM  AGIN  ARIES.  20? 

quantity  itself;  hence  \/^/  x  -v/— a  =  —a.  But  if  we  apply  the 
process  of  multiplying  the  quantities  under  the  radicals  we  have 
^—ax  ^J —a  =  V^'  =  +a,  as  well  as  —a.  What  then  is  the 
product  of -v/— ax 'v/— a?  Is  it  —a,  or  is  it  both  +a  and  —a? 
The  true  product  is  —a ;  and  the  explanation  is,  that  ^a^  is,  in 
:eneral^  +  a  or  —a.  But  when  we  know  what  factors  were  multi- 
plied together  to  produce  a'^,  and  the  nature  of  our  discussion  limits 
us  to  these,  the  sign  of  -y/a'  is  no  longer  ambiguous ;  it  is  the  same 
as  was  its  root. 


24:1,  Prop. — Every  unaginary  term  of  the  second  degree 
(and  in  fact  of  every  other  degree)  can  be  reduced  to  the  form 
mV — 1  in  which  m  is  not  imaginary,  m  may  he  rational 
or  surd. 

Demonstration. — Let  ^—x  represent  any  such  expression.  Then 
^/—x  —  ^x{—V)  =  '\/aj\/— 1,  which  is  the  required  form. 

242,  Scholium  3.  —  By  means  of  this  proposition  and  the 
property  noticed  in  Sch.  2,  the  multiplication  and  division  of 
imaginaries  is  effected . 

EXAMPLES. 

Ex.  1.  Multiply  W^-S  by  2\/^. 

Operation.  4  \/—S  =  4  \/3  x  V-l-  ^so  2  \/^  =  2  V^ 
X  V^-  Hence  4  ^—d  x  2  \/^  =  2  x  4  \/3  x  a/2  x  \/^ 
X  V^  =  8  \/6  X  \/^  X  -v/^  =  —S\/Q,  since  V— 1  x  \/—l 
=  -1. 

2.  Multiply  SV-^  by  4\/^.        Prod.,  -I^a/TsT 

3.  Multiply  V—^  by  V— p.  Prod.,  —xy, 

4.  Multiply  2\/^  by  3\/^.  Prod.,  -36. 

6,  Multiply  2V^y  5\/^,  SV^  together. 

j>rod.,   -60V4^V^,  or  -0Q>/^^, 


208  CALCULUS   OF   RADICALS. 


6.  Multiply  24H-\/-49  by  24-a/— 49. 

Suggestion. — We  have  here  the  product  of  the  sum  and  difference 
of  two  quantities  which  equals  the  difference  of  their  squares.    Hence 

(34  + ^1149)  X  (34-/v/^^^)  =  (24)'-(V^^^)'=576-(-49)=576 
+  49  =  625. 


7.  Divide  V— 16  by  V— 4. 

Operation.      ^/—lQ  =  4  \/—lj  and  V— 4  =  2  -v/— 1-      Hence 

.^_  4      2V-1 

Scholium. — A  superficial  view  of  the  case  might  lead  one  to  think 
that  the  quotient  w^as  ±2.     Thus,  noticing  that  the  radicals  are 

similar,  he  might  conclude  that  "^      —  =  y  ~—-  =  -y/I  =  ±2,  an 

y^—  4       '^    —  4 
incorrect  result. 

8.  Divide  6\/^  by  2V^.  4>t^o^.,  |a/3. 

2 

9.  Divide  —  V^  by  — eV^.  QuoL,  ^Vs. 

18 

10.  Divide  l  +  V^  by  1  — V^. 

Operation. — Writing  the  example  thus ^ •  and  multiply- 

1  — V— 1 

ing  both  terms  of  the  fraction  by  1  +  ^/^—i  there  results,  (see Ex.  6) 

l  +  2\/^-l       2^/^ 


l-(-l)  2 


=  V-i. 


The  example  may  also  be  performed  thus : 

1  +  V^   I  -  A//^"!  +  1 


1  +  y_i       ^31 

Since  1  divided  by  —  ^^  gives  y^,  as  —  y^xy'^ 


SYN'OPSIS. 


209 


H 

O 

fic    5 

I  -J 
a  I  o 

I  I 

CO 

u 

o 

0. 


SYNOPSIS    FOR    REVIEW. 

Power.— Degree  of.  J  g^H  l.-Powers  and  Roots  correlative. 


r.  ) 


Root.— Degree  of. 

A  r  1.  +  Integer.  |  ,, 
-|  2.  +  Fraction.  >■  . 
■  I  .'J.  -  Int.  obFr.  ) 


ScH.  2.— Square,  Cube,etc. 


Exponen 
or  Index 


Radical. 


1.   +  Integer. 
+  Fraction. 
-  Int.  or  Fr. 

('  j  Ratioual. 

Real.  i ,     ^, 

'  Irrational. 

Imaoinaby. 

I     How  INDICATED. 

I 


Cor.  Transferring  a  factor  in  a  frac- 
ion  from  one  term  to  another. 


\f 


1. 

+m. 

2. 

n 

8. 

-men 

_m 

n* 

Sim.  Rads.     rationalization.    To  afiect  witli  Exponents. 
Involution,     evolution.    Calculns  of  radicals. 
Prob.   I.     To  any  power.    Rule.     Dem.     Cor.— Signs. 


Prob.  2.     To  affect  with  Exponents.  Rule.  Dem.^ 


Cor.  1. — Terminates. 
Cor.  2.— Number  of  Terms. 
Cor.  3.— Coeflacients. 
Q)r.  4.— Exponents  in  ea.  term. 
Cor.  5,— RuLK. 
l^  Cot:  Q.-(a-b)'". 

Sch.— Signs. 

Cor.  1.— Same  as  (193). 

Cor.  2.— R.  of  Prod.  =  Prod. 

of  Roots. 
Cor.  3  — R.  of  Quot.  =  Quot 

of  Roots. 


Prob.  3.    Bmojcuii  Fosxula. 


Prob.  I. 


Roots  of  Perfect  Powers 
RuLB.    Dem. 


I  Sch.  1. 
Prob.   2.     Square  Root  of  Poly.    Rulb.    Dem.  <  8071.2. 

(  Sch.  3. 

Prob.  3 
Prob.  4 

Prob.  5 


Square  Root  of  Dec.  No.    Rule.    Dem. 
Cube  Root  of  Poly.    Rule.    Dem 


Sch.  1. 
Sch.  2. 

Sch.  1. 


Sch.  2, 
Cube  Root  of  Dec.  No.    Rule.    Dem 


•\ 


Sch.l. 
Sch.  2, 
Sch.  8. 


Prob.  6. 
Prob.  7. 


Any  Root  whose  index  composed  of  fiictors,  2  or  8. 
Any  Root.     Solution,    Gen^r^l  Scholiqm, 


210 


SYNOPSIS. 


CO 

< 

o 

o 
< 

< 
z 


€0 

O 

< 


z 
o 

o  i 

o 


SYNOPSIS    FOR    WEVIEW,— Continued. 

Prob.   I.  Remove  factor.    Rxjle.  j  Sch.  Simplest  form. 

Dem.  I  Cor.  Denomination  of  surd. 

Prob.   2.  Index  composite,  etc    Rule.    Dem. 

Prob.  3.  To  given  index.    Rule.    Dem. 

Prob.  4.  To  common  index.    Rule.    Dem. 

j    Prob.   5.  To  rationalize  Monomial  Denominator.    Rule,    Dem. 

Prob.   6.  To  rationalize  Binomial  Denominator.    Bxrus.    Dem. 

Prop.   I.  To  rationalize  any  binomial.    Dem. 

[^  Prop.   2.  To  rationalize  any  trinomial.    Dem. 

f  Prob.   I.      To  add  and  subtract.    Rule.    Dem. 
3  C  Prop.   I.       Product  of  roots  =  equal  root  of  product.    Dem. 
^  •<  Prop.   2.      Similar  Radicals,  how  multiplied.    Dem. 
S   '  Prob.  2.     To  multiply  Radicals.    Rule.    Dem. 


CO 
UJ 

£ 

< 
z 

< 

11 


Prop.     Quotient  of  roots  =  root  of  quotients. 
Prob.   3.      To  divide  Radicals.    Rule.    D&n. 


Involution  of  Radicals.    Rule,    Dem.    Sch. 
Evolution  of  Radicals.    Rule.    Dem. 


Prob.  4. 
Prob.  5. 

Prob.  6.  Va+  Vft  and  V  Va+  Vh,  etc. 

.  j  Sch.  1.— Called  Impossible. 
Definitions,  -j  ^^^   2._Reason  for  name. 

Prob.   I.  To  add  or  subtract.    Rule.    Dem. 

Prob.  2.  To  multiply.    Rule.    Dem. 

Prob.  3.  To  divide.    Rule.    Dem. 


Test  Questions. — By  what  must  numerator  and  denominator  of 
— ^    be  multiplied  to  reduce  it  to  a  simple  fraction  ?    Give  the 


various  significations  of  an  exponent.  Perform  the  operation 
VS  X  v^3,  and  explain  the  process.  Repeat  the  Binomial  Formula, 
and  by  means  of  it  expand  {l—x")^.    Demonstrate  the  rules  for 

square  and  cube  root.  Show  that ; • 

What  sign  is  given  to  a  square  root?  Why?  To  a  cube  root? 
Wliy  ?     What  is  the  value  of  a?"  ? 

[Note.— Here  ends  the  subject  of  Literal  Arithmetic.  The  student 
is  now  prepared  for  the  study  of  Algebra,  properly  so  called  ;  i.  e., 
The  Science  of  the  Equation.^ 


PART     II. 

LGEBRA. 


iSiliEciiiif] 


^ 


Sectjon  I. 


EQUATIONS  WITH    ONE    UNKNOWN    QUANTITY. 


DEPINITIONS. 


1.  An  liquation*  is  an  expression  in  mathematical 
symbols,  of  equality  between  two  numbers  or  sets  of 
numbers. 

Illustration,     ^x  —  2a''y  =  -    —-  is  an  equation  because  it  is 

an  expression  of  equality  between  Sx—2a^y  and  ^-—  • 


2,  Algebra  is  that  branch  of  Pure  Mathematics  which 
treats  of  the  nature  and  properties  of  the  Equation  and  of 

*  Do  not  pronounce  this  word  "  Equazion."    For  this  common  error  there  Is  no 
authority.    ''  Equashun  "  is  the  correct  pronunciation. 


212  SIMPLE    EQUATIONS. 

its  use  as  an  instrument  for  conducting    mathematical 
investigations. 

3,  The  First  Member  of  an  equation  is  the  part  on 
the  left  hand  of  the  sign  of  equality.  The  Second 
Member  is  the  part  on  the  right. 

4.  A  Numerical  Equation  is  one  in  which  the 
hnown  quantities  are  represented  hy  decimal  numbers ;  as 
l%x^—^x  =  48. 

5,  A  Literal  Equation  is  one  in  which  some  or  all 
of  the  known   quantities  are   represented   by  letters;   as 

,     ^,  4:00^  —  2 

ax—c-^Uy  =  —^ 

6.  The  Degree  of  an  Equation  is  determined  by 
the  highest  number  of  unknown  factors  occurring  in  any 
term. 

Illustration,  ax—lx^  =  c  +  ai^,  is  of  the  3rd  degree ;  a^x—4:x  =  12 
is  of  the  1st  degree ;  x^y^  =  18  is  of  the  4th  degree,  etc. 

7.  A  Simple  Equation  is  an  equation  of  the  first 
degree. 

Illustration.    y  =  ax-\-h  is  a  simple  equation,  as  also  is  — -— 

8.  A  Quadratic  Equation  is  an  equation  of  the 
second  degree. 

9,  A  Cubic  Equation  is  an  equation  of  the  third 
degree.    A  Biquadratic  is  one  of  the  fourth  degree. 

10,  Equations  above  the  second  degree  are  called 
Higher  Equations. 


WITH    ONE    UNKNOWN    QUANTITY.  218 

TRANSFORMATION  OP  EQUATIONS. 

11,  To  Transform  an  Equation  is  to  change  its 
form  without  destroying  the  equality  of  the  members. 

12.  There  are  four  principal  transformations  of  simple 
Equations  containing  one  unknown  quantity,  viz.:  Clearing 
of  Fractions,  Traiisposition,  Collecting  Terms,  and  Dividi7ig 
by  the  coefficient  of  the  unknown  quantity. 

IS,  These  transformations  are  based  upon  the  following 

AXIOMS. 

Axiom  1. — Any  operation  which  does  not  affect  the  value 
of  a  term  or  member ,  mag  be  performed  upon  that  term  or 
member  without  destroying  the  Equaiion. 

Axiom  2. — If  both  members  of  an  Equation  are  increased 
(yr  diminished  alike,  the  equality  is  not  destroyed. 


14,  Prob. — To  clear  an  Equation  of  Fpactions. 
Rule. — Multiply  each  member  hy  the  least  or  lowest 
common  multiple  of  all  the  denominators. 

Demonstration.— This  process  clears  the  Equation  of  fractions, 
since,  in  the  process  of  multiplying  any  particular  fractional  term, 
its  denominator  is  one  of  the  factors  of  the  L.  C.  M.  by  which  we 
are  multiplying;  hence  dropping  the  denominator  multiplies  by 
this  factor,  and  then  this  product  (the  numerator)  is  multiplied  by 
the  other  factor  of  the  L.  C.  M. 

This  process  does  not  destroy  the  Equation,  since  both  members 
are  increased  or  diminished  alike. 

EXAMPLES. 

Ex.  1.  Clear  the  equation  |4- 1+  |  =  -^^  of  fractions. 

Model  Solution.  6  is  the  L.  C.  M.  of  2,  3,  6,  and  3,  the  denomina- 
tors. Now,  it  is  evident  that,  if  the  first  member  of  this  equation  is 
equal  to  the  second,  6  times  the  first  member  is  equal  to  6  times  the 


214  SIMPLE    EQUATIOIn'S. 

second;  hence  I  can  multiply  each  member  by  6  and  not  destroy 
the  equality. 

Therefore  taking  the  equation 

X      X      «_2«  +  3 

2  "^  3  "^  6  ~  "~3~  ' 
and  multiplying  each  member  by  6,  I  have 

^x  +  2x  +  x  =  4a;  +  6. 

The  operation  is  performed  thus  :  2  times  ^  isa;;  and  3  times  a;  is 

2 

X  X 

3a?.     3  times  -  is  x ;  and  2  times  x  is  %x.     6  times  ~  is  x.     Hence   6 
o  6 

times  the  first  member  is  3.x -\-2x  +  x. 

3  times  — ^—  is  2a;  +  3;  and  2  times  2a; +  3  is  4a; +  6.     Hence  6 
o 

times  the  second  member  is  4a; +  6. 

Thus  the  denominators  have  all  been  caused  to  disappear  without 
destroying  the  equality  of  the  members  of  the  equation,  as  both 
have  been  increased  alike. 

Illustration. — An  equa- 
tion is  aptly  compared  to 
a  pair  of  scales  with  equal 
arms,  balanced  by  weights 
in  the  two  pans. 

Now,  if  the  weights  in 

the  scale  pans  balance  each 

other,  that  is,  are  equal,  and 

we  multiply  the  weights 

in  each  pan  by  6  (or  any  other  number),  the  balance  (equality)  will 

still  be  preserved.     Or,  if  we  increase  or  decrease  the  weights  in 

both  pans  equally,  the  balance  (equality)  will  not  be  destroyed. 

2.  Clear  —  —  ~^i  =  ^oi  fractions. 

Result,  8^— 15a; +  12  =  QQ. 

3.  Clear  10  +  ?^  =  ^-^^  ^^  fractions. 

Result,  100  + 4.x— 1^  =  b  +  bx—^Ox. 

.    ^,        x—l       x—2  dx—1       4:—x    „„ 

4.  Clear  -^ -{-x  = —--  +  -~  of  fractions. 

4  6  bo 


WITH    ONE    UNKNOWN    QUANTITY.  215 

Suggestion. — The  multiplier  is  6.  In  multiplying  the  second  term, 

«, 2 

,  it  should  be  borne  in  mind  that  the  —  sign  preceding  this 

3 

compound  term,  shows  that  the  term  as  a  whole  is  to  be  subtracted. 
Hence  when  this  term  is  written  without  any  mark  of  aggregation, 
its  signs  are  to  be  changed,  as  in  removing  terms  from  a  bracket 
preceded  by  a  minus  sign.  The  equation  cleared  of  fractions  is 
Sx—d—2x  +  4:-\-Qx=z  3a:— 1  +  8— 2a!.     Why  are  not  the  signs  changed 

in  the  last  term,  -— -  ?  Are  all  the  signs  changed  in  the  term  — — -  ? 
o  o 

Yes.     What  becomes  of  the  —  sign  before  this  fraction  in  the  given 

example  ?     It  is  dropped  after  the  operations  signified  by  it  have 

been  expressed  in  detail.     We  might  write  the  equation  cleared  of 

fractions  thus :  3j:— 3— (2a?— 4)  +  6a;  =  3a;— 1  +  8— 2a;,  the  term  2a;— 4 

being  still  taken  in  the  aggregate.     Now  removing  the  parenthesis 

(give  the  reason)  we  have  3a;— 3— 2a;4-4  +  6a;  =  3a;— 1  +  8— 2a;,  as 

above. 

Neglect  to  make  this  change   of  signs  is  one  op  the 
most  common  mistakes  of  beginners. 


5.  Clear 1 s  =  4:C r —  of  fractions. 

m       am      tw  a^m 

Result,  a^mx  +  amx  --dhi  z=z  ^a^cm^—^mx  +  mn. 


6.  Clear a~  =  «  ^^  fractions. 

Result,  3a;— 6a; +  18  =  2a. 

7.  Clear  ^ ^r-^  -|-  -,  =  1  —  a;  of  fractions. 

da        %ah       a^ 

Result,  4a5a;— 3«a;  +  3a  +  18§  =  Wb—Qa^hz. 

8.  Clear ^  —  x  A ^  =  ^^^  —  1  of  fractions. 

a—o  a-{-o      a-{-o 

Suggestion. — The  multiplier  is  a'— ft". 

Result,  ax-irhx—ahi-^tl^x-^i^a—U  =  (a—by—a^-^l/^. 


216  SIMPLE    EQUATIOITS. 

9.  Clear  = ^  of  fractions. 

x—c       x-\-2c 


ax-\-2aC'-bx—2bc  =  ax-\-hx—ac—hc, 

X 

=  3  -\- '  of  fractions. 

Za—b 

Result,  4:ax^2bx  =  W—lbab  +  Qb'^-\-ax—2bx. 


2iX  X 

10.  Clear -^  =  3  + of  fractions 

a— 2b  2a— b 


TRANSPOSITION. 

15.  Transposing  a  term  is  changing  it  from  one 
member  of  the  equation  to  the  other  without  destroying 
the  equality  of  the  members. 

16.  Prob. — To  transpose  a  term. 

Rule. — Dj^op  it  from  the  in  ember  in  which  it  stands 
and  insert  it  in  the  other  member  ivith  its  sign 
changed. 

Demonstration. — If  the  term  to  be  transposed  is  +,  dropping  it 
from  one  member  diminishes  that  member  by  the  amount  of  the 
term,  and  writing  it  with  the  —  sign  in  the  other  member,  takes 
its  amount  from  that  member ;  hence  both  members  are  diminished 
alike,  and  the  equality  is  not  destroyed.     (Repeat  Axiom  2.) 

2nd. — If  the  term  to  be  transposed  is  — ,  dropping  it  increases  the 
member  from  which  it  is  dropped,  and  writing  it  in  the  other 
member  with  the  +  signiwcreasesthatmember  by  the  same  amount; 
and  hence  the  equality  is  preserved.     (Repeat  Axiom  2.) 

EXAMPLES. 
Ex.  1.  Given  the  equation  3  + 2a;— 5  =  12— 4rr  to  trans- 
pose so  that  all  the  terms  containing  the  unknown  quantity, 
X,  shall  stand  in  the  first  member  and  the  known  terms  in 

the  second  member. 

Model  Solution. 
Operation.        3  +  2a:— 5  =  12— 4aj 
2x  +  4:x  =  12—3  +  5 
Explanation. — Dropping  3  from  the  first  member  diminishes  that 
member  by  3 ;  hence  to  preserve  the  equality,  I  subtract  3  from  the 


WITH    ONE    UNKNOWN    QUANTITY.  .      217 

second  member,    or  indicate  the  subtraction  by  writing  it  in  the 
second  member  with  the  —  sign.     Thus  the  term  3  is  transposed. 

Dropping  —5  from  the  fii*st  member  incrmses  that  member  by  5 ; 
and  hence  to  preserve  the  equality  I  add  5  to  the  second  member. 
Thus  the  term  —5  is  transposed. 

Dropping  —4a;  from  the  second  member  increases  that  member  by 
4x ;  hence  I  increase  the  first  member  by  adding  Ax  to  it,  and  thus 
preserve  the  equality. 

I  have  thus  arranged  the  terms  so  that  all  those  containing  the 
unknown  quantity  stand  in  the  first  member,  and  all  known  terms 
in  the  second  member ;  and  yet  I  have  preserved  the  equality. 

Illustration. — This  operation  can  be  illustrated  by  the  scales  on 
page  214.  Suppose  the  positive  terms  to  represent  weights  and  the 
negative  terms  some  forces  lifting  on  the  scale-pans.  [Since  the  + 
and  —  terms  represent  quantities  opposed  in  effects.]  Taking  the 
8  from  the  first  member  corresponds  to  taking  off  so  much  weight. 
This  can  be  compensated,  so  as  to  keep  the  scales  in  equilibrium,  by 
applying  a  lifting  power  of  0  to  the  other  side,  which  is  symbolized 
by  —3  on  that  side.  Again  taking  —5  from  the  first  member  is 
like  taking  away  a  lifting  power  of  5,  which  can  be  compensated  by 
putting  a  «?e/^A^  of  5  on  the  other  side  (  +  5).  In  like  manner  the 
transposition  of  any  term  can  be  illustrated.  In  fact  all  operations 
ujion  equations  can  he  illustrated  in  a  similar  way. 

In  the  following  examples  transpose  the  unknown  terms  to  the 
first  member  and  the  known  to  the  second. 

2.  Given  5x—12a-^3c—2a;=:4:X—2x-^4:a  to  transpose 
as  above. 

Result,  6x—2x—Ax-^2x  =  4:a-\-12aSc. 

3.  Given  100-f 4a;— 6  =  5a;  +  5— 30ir  to  transpose  the 
terms  as  above. 

Result,  Ax—bx-\-  30a;  =  5  —  100  +  6. 

.1    m  ,         15       3a;      ,^      7       1       11 

4.  Transpose  as  above  -^  — -^  +10—  -  =  -  —  —x-^x. 

T,      ,,   11  3a;       1       7       ,^       15 

Result,-x-x--  =  -^^-^-X^^-. 

10 


218  SIMPLE    EQUATIONS. 


5.  Transpose  as  above  Sab 4:ax-{-lSbx^  =  -~^—xK 

Eesult,  ISbx^  j^  x^  —  ^ax —  — Sab. 

5  c 


SOLUTION    OP    SIMPLE    EQUATIONS    WITH    ONE 
UNKNOWN    QUANTITY. 

17,  To  Solve  an  Equation  is  to  find  the  value  of  the 
unknown  quantity:  that  is,  to  find  what  value  it  must  have 
in  order  that  the  equation  be  true. 

Illustration. — In  the  equation  4a;— 2  =  2aj  +  4,  if  we  call  a*,  3,  the 
first  member  is  10,  and  as  the  second  is  also  10  for  this  value  of  «, 
the  equation  is  true.  But  if  we  try  any  other  number  than  3 
for  X,  we  shall  find  that  the  equation  will  not  be  true.  Thus 
trying  4  for  a?,  we  find  the  first  member  14  and  the  second  12 ;  and 
the  equation  is  not  true.  Again,  try  5.  The  first  member  becomes 
18  and  the  second  14,  and  the  equation  is  not  true. 

Let  the  student  see  if  he  can  ascertain  by  inspection  what  are  the 
values  of  x  in  the  following  : 

JB  +  3  =  3a7  +  1. 
2a;  =  30  —  X. 

Though  these  equations  are  very  simple,  it  is  probable  that  it  will 
take  the  student  some  time  to  guess  out  the  values  of  a;  which  make 
them  true. 

But,  if  it  is  so  difficult  to  hit  upon  just  the  value  of  x  which  is 
required  to  make  so  simple  an  equation  true,  the  task  would  be 
quite  hopeless  in  such  an  one  as 

3x-l  _  13-^  _  Z^  _  ll(a;  +  3) 
5  2  ~~  3  6        " 

Yet  we  have  a  very  simple  method  of  solving  any  such  equation 
so  as  to  tell  certainly  and  easily  what  the  value  of  a;  is.  This  pro- 
cess is  now  to  be  explained,  and  is  called  Solving  the  Equation^  or 
sometimes  the  Eesolution  of  the  Equation. 


WITH    ONE    UNKNOWN    QUANTITY.  219 

18,  An  equation  is  said  to  be  Satisfied  when  a  value 
is  given  to  tlio  unknown  quantity  which  makes  it  a  true 
equation :  i,  e.,  which  makes  its  members  equal.  It  is  said 
to  be  Destroyed  when  its  members  by  any  process  become 
unequal. 

W,  To  Verify  an  equation  is  to  substitute  the  sup- 
posed value  of  the  unknown  quantity  and  thus  see  if  it  sat- 
isfies the  equation. 


20.  Prob.  1.— To  solve  a  simple  equation  containing  one 
unknown   quantity. 

Rule. — /.  //  the  equation  contain  fractions,  clear 
it  of  them  by  Art  14. 

//.  Transpose  all  the  terms  involving  the  unknown 
quantity  to  the  first  memher,  and  the  known  terms  to 
the  second  member  by  Art.  15. 

///.  Unite  all  the  terms  containing  the  unknown 
quantity  into  one  by  addition,  and  put  the  second 
member  into  its  simplest  form. 

IV.  Divide  each  member  by  the  coefficient  of  the 
unknown  quantity. 

Demonstration. — The  first  step,  clearing  of  fractions,  does  not 
destroy  the  equation,  since  each  member  is  multiplied  by  the  same 
quantity  (Axiom  2). 

The  second  step  does  not  destroy  the  equation,  since  it  is  adding 
the  same  quantity  to  each  member,  or  subtracting  the  same  quantity 
from  each  member  (Axiom  2). 

The  third  step  does  not  destroy  the  equation,  since  it  does  not 
change  the  value  of  the  membei*s  (Axiom  1). 

The  fourth  step  does  not  destroy  the  equation,  since  it  is  dividing 
each  member  by  the  same  quantity,  and  thus  changes  the  members 
alike  (Axiom  2). 

Hence,  after  these  several  processes,  we  still  have  a  true  equation. 
But  now  the  first  member  is  simply  the  unknown  quantity,  and  the 
second  member  is  all  known.  Thus  we  have  what  the  unknown 
quantity  is  equal  to  ;  i.  e.,  its  value. 


220  SIMPLE    EQUATION'S. 

21*  Scholium  I. — It  must  he  fixed  in  the  pwpiVs  mind  that  he  can 
make  hut  two  classes  of  changes  upon  an  equation  without  destroying  it : 
viz,,  Such  as  do  not  affect  the  value  of  the  members,  or 
SUCH  AS  AFFECT  BOTH  MEMBERS  EQUALLY.  Every  Operation 
must  I>e  seen  to  conform  to  these  conditions. 

EXAMPLES. 
Ex.  1.  Solve  — ^  -\ h  o  =  — o^  y  a^d  verify  the 

/ii  O  o  o 

result. 

Model  Solution. 

x  +  1       3a;-4      1      603  +  7 
Operation.        (1) _  +  -^-  +  ^  =  -- , 

(3) 20x  +  20  +  24a;-33  +  5  =  30a?+35, 

(3) 20o?  +  24«-30aj  =  35-20  +  32-5, 

(4) 14«=42, 

(5) x=3. 

Explanation.— I  first  clear  equation  (1)  of  fractions  by  multiply- 
ing each  member  by  the  L.  C.  M.  of  its  denominators,  which  is  40. 

aj  + 1 
This  does  not  destroy  the  equation  (Axiom  2).     I  multiply  — ^r— 

2 

by  40  by  dropping  its  denominator  2,  thus  getting  a?+l,  as  the 
result  of  multiplying  by  2,  and  then  multiply  x  +  1  by  20,  getting 
20a;  +  20.  [In  like  manner  explain  the  entire  process  of  clearing  of 
fractions.] 

Having  cleared  the  equation  of  fractions  I  have  (2).  I  now  transpose 
the  terms  containing  x  to  the  first  member  and  the  known  terms  to 
the  second  member.  Thus,  dropping  SOo?  from  the  second  member 
diminishes  it  by  that  amount,  whence  to  preserve  the  equality  of  the 
members  I  subtract  30x  from  the  first  member,  i.  e.,  write  it  in  that 
member  with  its  sign  changed.  [In  like  manner  explain  the  transpo- 
sition of  each  term.] 

I  know  equation  (3)  to  be  true,  since  I  have  changed  both  mem- 
bers alike,  that  is,  have  added  to  and  subtracted  from  each  member 
the  same  quantities,  I  now  add  together  the  terms  of  the  first 
member,  which  does  not  affect  the  value  of  the  member,  and  have  14«. 
In  the  same  manner  uniting  the  terms  of  the  second  member  does 
not  alter  its  value  ;  hence  14a?  =  42,  Finally,  I  divide  each  mem- 
ber by  14  and  have  a;  =  3.  This  operation  does  not  destroy  the 
equation,  since  each  member  of  the  equality  14a;  —  42  is  divided 
by  the  same  number  (Axiom  2).     Hence  3  is  the  value  of  «, 


WITH    ONE    UNKNOWN    QUANTITY.  321 


Verification. 

I  will  now  see  by  actual  trial  if  3  does  satisfy  the  given  equa- 
tion.    Substituting  3  for  a:,  I  have 

3  +  1      3x3-4      1      6x34-7 

4      5      1      25 

2^5  +  8  =  ¥'"^ 

2  +  l+^  =  3|,or 

the  members  of  which  are  identical ;  and  the  value  of  a;  is  verified. 

2.  Solve  and  verify  — - —  =:  — - —      Result,  x  =z  1, 

,,     .«     ..  3  +  1      7  +  5        3      12      ,,^         .,  ,.    ^ 

Venflcation.     — ^r-  =  — --,  or  -  =  —  •    Whence  it  appears  that 

1  satisfies  the  equation. 

3.  Solve  and  verify  2x—-  =  — ^r 1 5— • 

Result,  a;  =  2. 

4.  Given  -^ — -  —  -=  =  — = ^  to  find  x. 

6  0  7  0 

Result,  X  =  4:, 
6.  Given  7a;  +  6— 3a;  =  564-2a;  to  find  x,  and  verify. 

6.  Given  ^-  +  6?/  =  ^^^  to  find  y. 

ff    n-        o.    .   3a;-ll       bx-b   ,   m-lx^    .    , 

7.  Given  21  H -r —  =  —5 1 ^^ —  to  find  x. 

Result,  a:  =  9. 

o    -,.         5(a:-5)    ,    .         284  -  a;  ,    ^    , 

8.  Given    ^  ^^      4-  6a;  =  — - —  to  find  x. 

20  5 


222  SIMPLE    EQUATIONS. 


9.  Given  25(lla:— 8)  =  18(12a;  +  2)  to  find  x, 

X. 

Result,  X  =  10. 


Q/v.  ^ 

10.  Given  —  +  2^+11  =  7  +  17  to  find  x 
5  4 


11.  Given  y-\ ^  =  — g--  to  find  y. 

Result,  y  =  34. 
Verification.     3f  +  ^^^  =  i^?^,  or 

which  is  a  true  equation,  since  20f-^3  is  6f. 

In  verifying  it  is  not  well  to  go  through  the  processes  of  clearing 
of  fractions,  transposition,  etc. ,  but  rather  keep  the  terms  as  distinct 
as  possible,  and  reduce  each  member  separately  to  a  form  so  simple 
that  they  can  be  seen  to  be  equal. 

y o  g     I    „ 

12.  Given  — ^ \-  3z  =  — \-  1,  U)  resolve  and  verify  as 

above.  Result,  z  =  1-^. 

Verification.     ^^        +  17  ~  ~~2      "^    '  ^^^^^^  reduces  ^0  —  — 
75_55       17         72_72 
■^  17  ""  17  "*"  17 '  °^  17  ~  17  ■ 

13.  Given  %  +  ^-i^  —  3  =:  ^~^^^  -  2^  to  resolve  and 

verify  as  above.  j?      u       _  ^-'■ 

Kesmt,  y  —  Tng* 

14    Given  ?(^zi)  +  ?(^^  -  M.^_^). 

14.  Lriven        5       +       15-3 

Result,  x  =  —' 


WITH    ONE    UNKNOWN    QUANTITY.  223 

15.  Find  the  value  of  a;  m  3a;  H z —  =  5  H ^r 

0  2 

Suggestion. — Having  cleared  this  equation  of  fractions,  transposed 
the  unknown  terms  to  the  first  member,  and  the  known  to  the 
second,  there  results  —  31a;  =  —147.  Hence  to  obtain  a  positive 
result  divide  each  member  by  —31,  and  x  =  7. 

22*  Cor.  1. — All  the  si^s  of  the  terms  of  both  members  of 
an  equation  can  be  changed  from  -f  to  — ,  or  vice  versa,  with- 
out destroying  the  equality,  since  this  is  equivalent  to  multiply- 
ing or  dividing  each  member  by  — 1. 

16.  Find  the  value  of  ^  in  ic  H ^-^  =  -^ 2. 

Result,  X  =  L 

97-7y  ,  6{y-l)       ^,    ,   3«/-ll  ,    .    . 

17.  Given  — ^  +  -^-^  =  21  -f-    ^  to  find  y. 

/i  o  lb 

Result,  y  =  9. 

18.  Find  the  value  of  ;z  in  1  +  ^±^  =  Sz  -^  ^-^.    (See 
'^  Result,  z  =  ~ 

19.  Find  the  value  of  x  in  —J-  +  -^  =  16 ^. 


Suggestion. — In  clearing  of  fractions  be  careful  to  notice  the  term 
—  •  As  this  term  is  to  be  subtracted,  when  the  sign  of  aggre- 
gation (the  line  between  the  terms  of  the  fraction)  is  dropped,  the 
signs  of  the  separate  terms  must  be  changed.  The  equation  when 
cleared  is  6aJ4-6-|-4a?-|-8  =  193— 3aj— 9;  and  «  =  13.     It  is  well 

aJ4-  3 
for  the  pupil  to  explain  such  cases  thus  :     Having  multiplied  — — - 

by  the  L.  C.  D.  13, 1  have  3a;  +  9 ;  but  this  is  to  be  subtracted,  hence 
I  annex  it  with  its  signs  changed,  as  subtraction  is  performed 
by.--.. 


2M  SIMPLE    EQUATIONS. 

20.  What  value  for  z  satisfies  z  -\ ^r—  =12 ^  ? 

21.  Find  the  value   of  z  which   satisfies  the  equation 
^"^  4  ^-3  12 


22.  Given  — r^ — —  =  —. 2f ,  to  find  x. 

14  21  4  * 


Separating  the  fractions  into  parts,  and  reducing,  j3^a;+|^—/xa; 
+i  =  i«-l-2f.  Whence  (^8^-^\-i)a;  =  _^-  i-3|,  or  — ^o; 
=  — If,  or  |a!  =  5,  and  x  =  35. 

23.  Given  ^(7x  +  9)-{-^{2^9-9x)  =  i(9x-ld)-}-\{Sx 
+  1),  to  find  X. 

24.  Solve  i(a;+l)+i{^  +  2)  =  lQ-i{x-\-S). 

25.  Solve  j(3.2;-3)-i(3a;-4)  ==  5^-^(27+ 4a;). 

26.  Sol\e  i{S—x)-\-x—li  =  i(x-^6)—^x, 

27.  Given  — — 6x  =  6,  to  find  x, 

a 

Operation. — Multiplying  each  member  by  awe  have  2a  +  x—5aa 
=  6a.     Transposing,  x—5ax  =  3a,  or  (1— 5a)a:  =  3a. 

[Note.— It  is  the  common  experience  of  pupils  that  they  continue 
to  find  difficulty  with  Literal  Equations  even  after  they  are  quite 
familiar  with  Numerical  ones,  such  as  the  26  above  given.  This 
difficulty  must  be  overcome.  No  one  has  caught  the  true  spirit  of 
mathematical  reasoning  till  the  literal  notation  is  seen  and  felt  to  be 
more  simple  than  the  decimal.] 

28.  What  value  for  x  satisfies  ax-\-h  =  cx-\-d?  Verify  it. 

d-b 
Ans.,  X  = 


WITH    ONE    UNKNOWN    QUANTITY.  225 

Verification.— Substituting  the  value  of  x  for  x  we  have  a-^^ 

a—c 

+  6  =  <5 ^J.    Performing   the  multiplications   indicated,  and 

ci — c 

reducing  each   member  to  an  improper  fraction,   this    becomes, 

ad—ab+ab—bc      cd—hc  +  aJ—cd      ...  ,       ,        ^  ,    . 

=  — Now  —ah  and  +ao destroy  each 

a—c  a—c 

other  in  the  first  member,  and  +cd  and  — cd  in  the  second.     Hence 

we  have =  — ,  which  is  evidently  true. 

a—o         a—c  '' 

29.  Given  ax -\- i^  =  bx-\-a^  to  find  the  value  of  x. 

Result,  X  =  a-\-b, 

30.  Given  057-1-771= 5a; -hn  to  find  a;.    Result,  x= ^» 

a — 0 

31.  Given  to-}- 2a;— a  =  3a;— 2c  to  find  x. 

Result,  X  =z  -J — - . 
0  —  1 

X       1 

32.  What  value  for  x  satisfies  ax-\-b  =  — |-t? 

a     0 

33.  Solve  — —  =  -^^^^  and  verify  the  result. 

771  c  *' 

34.  Find  X  from  ax aV^  •=ihx  A 

a  2a 

-^±i?.  Result  x-'^^^-^. 

4  4a— 35 

35.  Solve  ?=f  +  ?Z:*  +  ?=£  =  x-{a^+c)^ 

oca  abc 

Result  ^_^^c-\-al^-\-h(?-a-l)^c 
ac-\-ab-\-oc—\ 

36.  Given  -  -| —  =  c  to  find  x.       Result,  x  = 


a      x  ac — 1 


226  SIMPLE    EQUATIONS. 

««    ^.        x—a      2x—db      a—x      ^.     .  ..,  ,    „    , 
37.  Given  —5 h ^7-  =  lOo' + 11 J  to  find  a;. 

O  O  /« 

Result,  X  =  26a  +  24:b. 


38.  Given be  =  m  -^  -  to  find  a;. 


Result,  X  =  7 — : — • 
bc-\-m 

39.  Given  -- 1  +  Sab  =  — to  find  a;. 

a  c 

„      _,  ac(l— 3a5) 

Result,  X  =  — ^ r-^' 

c—ad 

40.  What  IS  the  value  of  x  m  — \-  - — --5-  =  — ^ — 

9  bo;  4-0  o 

23»  Scholium  3. — It  is  not  always  expedient  to  perform  the 
several  transformations  in  the  same  order  as  given  in  the  rule.  The 
process  may  often  be  much  shortened  by  the  exercise  of  a  little 
ingenuity.  The  ultimate  object  is  to  so  transform  the  equation  that 
the  unknown  quantity  will  stand  alone  in  the  first  member. 

An  expeditious  method  of  solving  the  last  example  is  as  follows: 
Multiplying  by  9,  we  have  6a!  +  7  +  — —  =  62:  +  12.     (The  term 

tj~, -J  o 

is  multiplied  by  3  by  dividing  the  denominator,  and  by  the 

other  factor  of  9,  by  multiplying  the  numerator.)     Now  dropping 

2\x 39 

Qx  from   each   member  and  subtracting   7,  we  get   — —  =  5. 

Whence  a;  =  4. 

.,     ^.         4a;  +  3    ,    7a:-29       80:4- 19,     n    -, 

41.  Given  —- h =  -—- —  to  find  x. 

9  52^—12  18 

Suggestion. — Multiply  numerator  and  denominator  of  the  term 
■^~-  by  2,  transpose  and  unite  it  to  the  second  member,  and  there 

results 3—  =  3-?:  •    Whence  cc  =  6. 

5aj— 13      18 


WITH    ONE    UNKNOWJf    QUANTITY.  327 

.„    _.        Qx+5   ,   8a;-7       36x+15       lOJ  ,    „    , 

42.  Gmn  — ^-  +  ^-^^  -^^  +  ^  to  find  x. 

43.  Given  «4tl  _  |^2  ^  3^ti  to  find  ^. 

15  7ic— 6  5 


44.  Given  —irz = tt;  =  — l —  to  find  x.  and  verify 

15  7a;— 16  5  '  ^ 

the  result.  Result,  x  =  —2. 

9 

—12  +  1        —4—4        —4—1  11       4 

Verification.    -^ ___  =  ___, or- j^ --  = -1, 

15         , 

45.  Given  -^ — i-J  H —  =  8  to  find  x,  and  verify 

the  value. 


46.  Solve  for  a;  the  equation  -  + 


a       h—a       h-\-a 

An  elegant  solution  of  this  is  obtained  by  first  uniting  the  two 

terms  of  the  first  member,  which  makes  the  equation  = — - 

a(b-a) 

aha         --7-  a^G)—a) 

=  i ,  or  -— — -  X  —  -= Whence  x  =  ^^ — { • 

6+a         a{b—a)  O+a  b(b-\-a) 

Suggestion. — Performing  the  multiplications  and  transposing,  we 

I         X      X      X      a       a        a  6  2„^  ,.  .,. 

^^'^'  2  -  3  +  r ^  6  -  12  "■  20 '  ^'  12''  =  IS"^-     ^'°'"   ^'^'^'°S 

u      5         -  8 

by  —  we  have  x  =  —a. 

48.  Solve  "{-'.=  '-  ^-y-  RuuU,  y  =  ^. 

0       a      a        c  ^       ad 


228 

SIMPLE 

EQUATIONS. 

49. 

Solve 

2x 
a^2b 

=  3  + 

X 

2a-b 

• 

Result,  X  - 

=  2a- 

-5i  +  ^. 
a 

Suggestion. — Transpose  - — ^  to  the  first  member  and  unite  the 
terms,  giving  -^-rz^^^-^.^  =  3.     .'.  x  =  2«-5&+-. 

50.  Solve  3x  —  a  =  X —  • 

o 

ei     o  1       x(a—b)  ,   ab      x 

51    Solve  -^— ^  ~  ^  "^  T  ~  3' 

ax—hx      X            ab        3a— 3Z>  +  2         ia+ab 
Suggestion.    — g— +  ^  =  «+ -;|,  or x=—^—.    .-. 

_4a  +  a5  6         _    3a(4  +  5) 


3a-3&  +  2      6(a-5)+4 

52.  Solve  ^-«  +  ?^±^  =  ^-^- 

„      ,,  8^— 4ac  +  «5c 

Result,  y=     ^^^^^_^^    » 

53.  Solve  |(a;-a)— j(2a;-35)-^(«-a;)  =  ^•-fa. 

Result,  x=.-rb. 
4 

54.  If  (^-%  _  (P±pl  +  3(«5  -  %by)  =  Sab,     show 


that  2/  =  0. 


55.  Solve  ^^  =  «.  +  f .  iJ..««,  X  =  t 

bx  0  ^ 

Suggestion.— First  divide  each  member  by  a,  and  then  write  the 
first  member  in  two  parts,  reducing  each  fraction  to  its  lowest  terms. 

Drop  zr  from  each  member  and  -  =  c. 


WITH    ONE    UNKNOWX    QUANTITY.  229 

56.  Solve  ^^  +  -  =  ..  +  1. 

Suggestion. — An  elegant  solution  of  this  is  obtained  by  no- 
ticinff    that    ^4; 7-  =  -; — --  =  x  +  1—- — -  ;  whence  the  equa- 

tion  becomes  a?+l— .r 1+5-  =  «+!•     Dropping  a:  +  l  from  each 

member,  transposing  and  changing  signs,   ^ =  —  ,  or  Sx  = 

2x+l.    .:  x=l. 

67.  Solve  -  H -:  =  -^—' 

x+2      11  3         1         ,         _ 

Suggestion.    ^  =  3  +  -.    .'.  ^^^  =  -,  and  ^  =  9. 

_^    ^  ,      6a;+13        3a;+5        2a;        ,        .. 
58.  Solve  -^^  -  --^  =  --,  and  verify. 


Suggestion. — First  destroy  the  term  —  • 

e#^     n  1       x—1        x  —  2        x  —  5        x—6 

69.  Solve  ^ ^  = 2. ji,' 

x—2      x—3      x^6      x—7 

Suggestion. — Reducing  each  term  to  a  mixed  number  we  have 

1  +  — --1 x  =  l  +  —  „-l ^.       Whence    — 

x—2  x—d  x—Q  x—1  x—2      x—S 

=  — .     Reducing  the  terms  in  each  member  separately  to 

1  -I 

common  denominators  and  adding,  we  get 


{x-2)(x-d)  (a;-6)(ir-7)' 
Whence  (x-2)(«-3)  =  (x-Q)  (x-l),  or  a;'-13a!  +  43  =  x^-5x  +  Q, 
and  x  =  4^. 

24,  Scholium  4. — It  often  happens  that  an  equation  which 
involves  the  second  or  even  higher  powers  of  the  unknown  quantity 
is  still,  virtually,  a  simple  equation,  since  these  terms  destroy  each 
other  in  the  reduction. 


230  SIM1>LE    :EQtATlOl!fS. 

60.  What  value  for  x  satisfies  S—x—2{x^l){x-^2) 
=  (x-3)  (6-2x)? 

Am.,  X  =  14. 

61.  Solve  the  equation  (x—6){x—2)  —  (x—6)(2x-5) 

Ans.,  x  =  2^. 

62.  Solve  ^  -  ?  (3.-4)  +  (I^^U^^^zH  =.  .. 

—  -- ,  and  verify. 

63.  Solve   (rc+ 1)  (a;- 1)  -  (^+5)  (x^3)  +  |  =  0,   and 
verify  the  result. 

64.  Solve  (a+x)  (h+x)  =  (c-\-x)  {d-{-x). 


Result,  X  = 


a-\-h—c—d 


65.  Solve  —^—  = h 


x—c      x—a      x—b 


Result   x  -  (^^J^^rh-^c)  ■ 
Result,  X  -  _-j-^^____. 


66.  Solve  [a  +  x)  {h  +  x)-a(h-^c)  =  y  +  x\ 

Suggestion. — Perform  the  multiplication  indicated  in  the  first 

member,  and  write  the  terms  without  clearing  of  fractions;  this 

,        XX  ^^^      ixn        ■    ,      Tx        0^0  + dbc      (a  +  I})ac 

gives  (a  +  d)x  —  ac=z  —'     Whence  {a+l>)x  =  — j^ —  =  ^—-^-. 

Dividing  by  (a+b)  we  have  ^  =  ^- 


SIMPLE    EQUATIONS     CONTAINING    KADICALS. 

2S,  Many  equations  containing  radicals  which  involve 
the  unknown  quantity,  though  not  primarily  appearing  as 
simple  equations,  become  so  after  being  freed  of  such 
radicals. 


IKVOLVING    RADICALS.  331 

26.  Prob.  2.— To  free  an  equation  of  Radicals. 

Rule. — /.  Transpose  the  terms  so  that  the  radical, 
if  there  is  bat  one,  or  the  more  complex  radical,  if 
there  are  several,  shall  constitute  one  member. 

II.  Involve  each  member  of  the  equalion  to  a  power 
of  the  same  degree  as  the  radical. 

III.  If  a  radical  still  remains  repeat  the  process, 
being  careful  to  Tceep  the  members  in  the  most  con- 
densed form  and  lowest  terms. 

Demonstration.—  That  this  process  frees  the  equation  of  the 
radical  which  constitutes  one  of  its  members  is  evident  from  the  fact 
that  a  radical  quantity  is  involved  to  a  power  of  the  same  degree 
as  its  indicated  root  by  dropping  the  root  sign. 

That  the  process  does  not  destroy  the  equality  of  the  members  is 
evident  from  the  fact  that  the  like  powers  of  equal  quantities  are 
equal.     Both  members  are  increased  or  decreased  alike. 

EXAMPLES. 


Ex.  1.  Find  the  value  of  x  in  V'^x-^-W  =  12. 
Model  Solution. 


Operation.        V^x  +  16  =  13. 
4a?  +  16  =  144. 
4x  =  128. 
a;  =  33. 

Explanation. — I  first  square  each  member  of  the  equation.  The 
first  member,  y^^c+lQ,  is  squared  by  dropping  its  radical  sign, 
since  the  square  of  a  square  root  is  the  number  itself  The  square 
of  the  second  member,  12,  is  144.  This  process  is  equivalent  to 
multiplying  the  first  member  by  \/ix  +  lQ  and  the  second  by  13, 
hence  as  \/4:x+16  is  equal  to  12,  both  members  have  been  increased 
alike.  [The  equation  being  freed  from  radicals  the  explanation 
becomes  the  same  as  before.] 


2.  Solve  V5x-\-6  =  4,  and  verify. 


232  SIMPLE    EQUATIONS. 


3.  Solye  VlO^  +  3  =  7.  Result,  xz=i~ 

o 


4.  Solve  V'2a;-f  3  +  4  =  7,  and  veiify. 

Suggestion. — First  transpose  the  4  and  unite  it  with  the  7; 
otherwise,  squaring  will  not  free  the  equation  of  radicals. 


6.  Solve  8+ v3a;+6  =  14,  and  verify. 


6.  Solve  and  verify  3a/2^T6  +  3  =  15. 

Result,  X  =  6. 

7.  Solve  Vax  +  2ah—a  =  h.  Result,  x  = 

a 

8.  Solve  Vx  +  x^—x^:^  =  0.  Result,  x  =  -> 


9.  Solve  «ic-|-a\/2aa;+a?*  =  a5.     Result,  x 


2(a  +  5) 

Suggestion. — Before  squaring  put  the  equation  in  the  form 

/\/'^ax  +  x  =  ft— 05. 

10.  Given  ^/l'^-\-y—^/y  =  2  to  find  the  value  of  y. 

Result,  y  =  4:. 

Queries. — If  each  member  is  squared  as  the  equation  stands, 
will  the  equation  be  freed  from  radicals  ?  Is  the  first  member  a 
binomial  or  a  trinomial  ?  What  is  its  square  ?  Which  will  give  the 
most  simple  result,  to  square  it  as  it  stands  or  to  transpose  one  of 
the  radicals  ?  Which  one  is  it  best  to  transpose?  Will  once  squar- 
ing free  it  from  radicals  ? 

11.  Given  Vx—16  =  S—\/x  to  solve  and  verify. 

Suggestion. — Solve  this  and  the  five  following  like  the  preceding 
by  squaring  twice. 


INVOLVING    RADICALS.  ^33 

12.  Solve  Va  +  a;+V«^  =  V«^. 

Result,  X  =  -5--—:  • 

13.  Solve  Vx—a  =  Vx—^Va, 

Result,  X  =  — r-« 
lb 


14.  Solve  V^xVx-\-2  =  V5X  +  2, 

Result,  ^  =  KK- 


16.  Solve  a  +  x  =  '^ d'^x^/W-^:i^, 

Result,  x  =  —J —  • 

16.  Solve  2VF+x—V^+x  =  \/x. 

Result,  x 


{h-al 
2a— h 


Query. — Why  is  it  best  to  transpose  one  of  the  terms  of  the  first 
member  to  the  second  before  squaring  ? 

17.  Given  x-\-^/a—x  =  --p=-  to  find  x. 

Va—x 

Scholium  I. — It  is  frequently  best  to  clear  the  equation  of  frac- 
tions first,  even  when  it  involves  radicals,  especially  if  the  denomi- 
nator is  of  such  a  form  as  will  not  make  the  equation  complicated. 

In  this  case,  clearing  of  fractions  we  have  x\/a—x  +  a—x  =  a. 
Whence  x\/a—x  =  x,  or  ^Ja—x  —  1.     Finally  x  =  a— 1. 

18.  Solve  — --  = 

yx  ^ 

Suggestion. — The  pupil  should  exercise  his  ingenuity  in  seeing 
in  what  difierent  ways  he  can  solve  such  examples,  and  notice  the 
most  elegant  methods.     For  example,  compare  the  following : 

1st  Method. — Clearing  of  fractions  x^—ax^  —  x.    Dividing  by  a;, 
,  1 

1— a 


234  SIMPLE    EQtJATlO:frS. 

%nd  Method. — Multiplying  each  member  by  y'ccwe  have  x—ax=\ 
as  before. 

^rd  Method. — Dividing  numerator  and  denominator  of  the  second 

member  by  ^/x^  we  have zz  =  — ^  •    Whence  multiplying  each 

^yx        ^x 

member  by  y^«,  x—ax  =  1,  etc. 

Mh  Method. — Divide  the  numerator  of  the  first  member  by  its 
denominator  and  ^Jx—a^Jx  =  ^^.   Dividing  each  member  by  'y/a;, 

1— a  =  - ,  or  «  = 

19.  Solve  — -= —  =  —^-z= Result,  — 

Vax-^h      Svax-^6b  ^ 

Suggestion. — The  more  elegant  solution  of  this  is  to  reduce  each 

member  to  a  mixed  number,  obtaining  1 := —  =  1 = . 

\/ax  +  l)  d^ax+5b 

2                    7 
Dropping  the  1,  and  dividing  by  —5,  — z= —  = = •   Clear- 

y^ax  +  b      S^ax  +  5b 

ing  of  fractions  Q\/ax  +  l()b  =  l^ax  +  lb.     .'.  ^Jax  =  Zb,  etc. 
In  a  similar  manner  solve  the  two  following : 

20.  Solve  -—-^-  =  -^^—-L —  Result,  x  =  4:, 

Vx-{-4:         Vx-\-6 

21.  Solve  -^^-—^ —  =  —=^ —      Result,  x  =  | -,  )• 

Vx  +  b        Vx  +  Sb  ^^-^^ 

22.  Solve  -^^-  r=  1  4-  ^^f~^.        Result,  a;  =  3. 

\/dx-\-l  ^ 

Suggestion.     — =iz —  =  -v/Saj— 1. 
^%x+l 

23.  Solve  -^^P^  =  Wx-^+  ^• 

Va:  +  2  ^ 


INVOLVING    RADICALS.  2^5 

Suggestion. — In  this  and  the  following  use  the  same  expedient  as 
in  the  22nd.    Thus  ^x—2=5^x—S  +  §\/x^  or  5^^x=Q,  ox  \/x  = 
13        .      ^1^ 
11'       '•'^"121* 

24.  Solve  -— —  = 5—  +  Wa. 

VX  H-  V  «  ^ 

Suggestion. — We    have    v^— V<*  =  i  V*+lf  V^»    or    \^/x 
—  2f /y/a.     .-.  X  =  16a. 

25.  Solve  — = =  c  +  -^^ 


1/  c2   \2 

Result,  X  =  -(b-\ )  • 

a\      c—lj 


26.  Solve —=r^r =Vm.    Result,  x  =  - — 

ya-{-x—Va—x  -• + 


m 


Suggestion. — Rationalize    the    denominator    (224),   and    after 
reducing  ^/a^—x*  =  x^/m—a. 


27.  Solve  ^« ±^  =  ,.         Result,  x  =  "-^l 

28.  Solve  — ^ =  9.  Result,  x  = -- 

^4x-\-l-2Vx  9 


29.  Solve  Vx+V^-  Vx^Vx  =  l\ /— 


Va 


Suggestion. — Multiply   each    member  by  Vx-\-\/x,  and  after 


reduction  y^x'—x  =  x—^\/x.    Squaring,   transposing,  and    com- 
25 
16' 


/-I      5  25 

bmrng,  ya?"  =  ^^-    .'.  x  = 


30.  Solve  V^/x-^^  -  V^/is  =  V^l 


Result,  a;  =  9. 


236  SIMPLE    EQtJATTOKS. 

31.  Solve  a/4^  +  J'^  =  \'/^- 

V  a+x       V  a—x       V  «  — ^ 

Result  X  =  -\ - 

h—c 

32.  Solve  -  +  -  =  V  -  4-  \/-^^  +  -4-        ^  =  ^«. 

33.  Solve  .15a:  +  1.575— .875a;  =  .0625ic.  x  =  2, 

1  Q/y. AK 

34.  Solve  1.2a:-         ,  —  =  Ax+%.^.  x  =:  20. 

.5 

lyo^ AK 

35.  Solve  4.8a;-  '     \   —  =  1.6a;+8.9.  x  =  6, 

.0 

36.  Solve  2.3a;-  ^=^?  =  3.2-  |. 

37.  Solvea;-.24=       ^^  •  -«  a;  =  4.259  +  . 

./ii. 


SUMMARY    OP    PRACTICAL    SUGGESTIONS. 

27. — In  attempting  to  solve  a  simple  equation  which 
does  not  contain  radicals,  consider, 

1.  Whether  it  is  best  to  clear  of  fractions  first. 

2.  Look  out  for  compound  negative  terms. 

3.  If  the  numerators  are  polynomials  and  the  denomina- 
tors monomials,  it  is  often  better  to  separate  the  fractions 
into  parts.     {Bx.  22,  p.  224  is  an  illustration.) 

4.  It  is  often  expedient,  when  some  of  the  denominators 
are  monomial  or  simple,  and  others  polynomial  or  more 
complex,  to  clear  of  the  most  simple  first,  and  after  each 
step  see  that  by  transposition,  uniting  terms,  etc.,  the  equation 
is  kept  in  as  simple  a  form  as  possible.     (See  Fxs.  40  to  44.) 

5.  It  is  sometimes  best  to  transpose  and  unite  some  of 
the  terms  before  clearing  of  fractions.     (See  Exs.  46  to  54.) 


APPLICATIONS.  237 

6.  Be  constantly  on  the  lookout  for  a  factor  which  can  be 
canceled  from  each  member  of  the  equation  {Ex,  55),  or  for 
terms  which  destroy  each  other  (Exs.  60  to  %Q), 

7.  It  sometimes  happens  that  by  reducing  fractions  to 
mixed  numbers  the  terms  will  unite  or  destroy  each  other, 
especially  when  there  are  several  polynomial  denominators. 
(See  Ex.  59.) 

28.  When  the  Equation  contains  Radicals,  con- 
sider, 

1.  If  there  is  but  o??e  radical,  by  causing  it  to  constitute 
one  member  and  the  rational  terms  the  other,  the  equation 
can  be  freed  by  involving  each  member  to  the  power  denoted 
by  the  index  of  the  radical.     (See  Exs.  1  to  9.) 

2.  If  there  are  two  radicals  and  other  terms,  make  the  more 
complex  radical  constitute  one  member,  alone,  before  squaring. 
Such  cases  usually  require  two  involutions.  (See  Exs.  10  to  16.) 

3.  If  there  is  a  radical  denominator  and  radicals  of  a 
similar  form  in  the  numerators  or  constituting  other  terms, 
it  may  be  best  to  clear  of  fractions  first,  either  in  whole  or 
part.     {^qqExs.  17,  18,  29.) 

4.  Frequently  a  fraction  may  be  reduced  to  a  whole  or 
mixed  number  with  advantage.     (See  Exs.  19  to  25.) 

6.  It  is  sometimes  best  to  rationalize  a  radical  denomina- 
tor.    (See  Exs.  26  to  28.) 

6.  Always  watch  for  an  opportunity  to  cancel  a  factor, 
or  drop  terms  which  destroy  each  other. 


APPLICATIONS. 


29.  According  to  the  definition  (3),  Algebra  treats  of, 
Ist,  The  nature  and  properties  of  the  Equation ;  and  2nd, 
the  method  of  using  it  as  an  instrument  for  mathematical 
investigation. 

The  Simple  Equation  has  been  so  thoroughly  discussed 


238  SIMPLE    EQUATIONS.  ,      * 

in  the  preceding  part  of  the  section,  that  the  pupil  will  now 
be  able  to  use  it  in  solving  problems. 

30.  The  Algebraic  Solution  of  a  problem  consists 
of  two  parts : 

1st.  The  Statement,  which  consists  in  expressing  by 
one  or  more  equations  the  conditions  of  the  problem. 

2nd.  The  Solution  of  these  equations  so  as  to  find 
the  values  of  the  unknown  quantities  in  known  ones.  This 
process  has  been  explained,  in  the  case  of  Simple  Equa- 
tions, in  the  preceding  articles. 

31.  The  Statement  of  a  problem  requires  some  knowl- 
edge of  the  subject  about  which  the  question  is  asked. 
Often  it  requires  a  great  deal  of  this  kind  of  knowledge  in 
order  to  '^ state  a  problem."  This  is  not  Algebra;  but  it 
is  knowledge  which  it  is  more  or  less  important  to  have 
according  to  the  nature  of  the  subject. 

32.  Directions  to  guide  the  student  in  the  Statement 
of  ProUems : 

1st.  Study  the  meaning  of  the  problem,  so  that,  if  you  had  the 
answer  given,  you  could  iwote  it,  noticing  carefully  just  what  opera- 
tions you  would  have  to  perform  upon  the  answer  in  proving.  This 
is  called,  Discovering  the  relations  l)etween  the  quantities  involved. 

2nd.  Represent  the  unknown  (required)  quantities  (the  answer) 
by  some  one  or  more  of  the  final  letters  of  the  alphabet,  as  x,  y,  0, 
or  wj,  and  the  known  quantities  by  the  other  letters,  or  numbers, 
given  in  the  problem. 

3rd.  Lastly,  by  combining  the  quantities  involved,  toth  known 
and  imJcnown,  according  to  the  conditions  given  in  the  problem  (as 
you  would  to  prove  it,  if  the  answer  were  known)  express  these 
relations  in  the  form  of  an  equation. 

PROBLEMS. 

Ex.  1.  What  number  is  that  to  the  double  of  which  if  18 
be  added,  the  sum  is  82  ? 


APPLICATIONS.  Ji39 


Model  Solution. 

Statement. — Let  x  represent  the  unknown  number  sought.  Then 
the  problem  says  that  double  this  number,  that  is  2x,  with  18  added, 
that  is  2ic  + 18,  is  82.     Hence  2a!+ 18  =  82  is  the  statement. 

Resolution  of  the  Equation.— [With  this  the  pupil  is  familiar.] 
Solving  2a;+18  =  82,  we  find  x  =  32.  32  is,  therefore,  the  number 
sought. 

Verlflcatlon.     2  x  32  + 18  =  82. 

[Note. — In  this  example  there  are  three  members  involved,  the 
18,  82,  and  the  one  to  be  found,  which  we  call  x.  The  relations 
between  these  numbers  are  explicitly  stated  in  the  problem,  and  the 
statement  is  easily  made.  This  is  not  always  so.  These  relations 
are  often  the  most  difficult  thing  to  discover,  and  their  discovery 
requires  a  knowledge  of  other  subjects  than  Algebra.  Suppose  the 
problem  to  be :  Given  three  masses  of  metal  of  equal  weight,  one  of 
pure  gold,  one  of  pure  silver,  and  one  part  gold  and  part  silver. 
When  they  are  immersed  in  water,  the  gold  displaces  5  ounces,  the 
silver  9  ounces,  and  the  compound  6  ounces.  What  part  of  the  last 
was  gold  and  what  part  silver?  Now  in  this  problem  the  relations 
between  the  quantities  are  not  explicitly  stated.  Yet  by  a  knowl- 
edge of  Natural  Philosophy  and  a  little  of  Memuration,  they  can  be 
found  out,  and  the  statement  of  the  problem  made  in  an  equation.] 

2.  What  number  is  that,  to  the  double  of  which  if  44  be 
added,  the  sum  is  4  times  the  required  number  ? 

Suggestion. — Suppose  I  guess  that  the  number  sought  is  30,  how 
will  you  tell  whether  I  am  right  or  not?  You  say:  "Double  30 
and  add  44  to  it,  and,  if  you  are  right,  the  sum  will  equal  4  times 
30."  But  2x30  +  44  does  not  =4x30;  so  I  am  wrong.  Now, 
call  the  number  sought  x,  and  use  it  as  30  was  used  in  proving  that 
it  is  not  the  answer,  and  the  statement,  2x-\-A4  =  4x  is  obtained. 
Whence  x  =  22.     Verify  it. 

3.  What  number  is  that,  the  double  of  which  exceeds  its 
half  by  6  ?  Ans.,  4. 


240  SIMPLE     EQL'ATIOKS. 

4.  The  property  of  two  persons  is  $16000,  and  one  owns 
three  times  as  much  as  the  other.     How  much  has  each  ? 

Statement.  Let  x  =  the  less  amount. 

Then  3a;  =  the  greater  amount. 
And  3a?  +  a;  =  16000. 

.'.  x  =  4000  and  3a;  =  13000. 

5.  A  man  being  asked  his  age  replied  :  "  If  to  my  age 
you  add  its  half,  and  third,  and  then  deduct  10,  the  result 
is  100."    What  was  his  age  ?  Ans.,  60. 

6.  After  paying  \  of  a  bill  and  \  of  it,  $92  still  remained 
due.     What  was  the  bill  at  first  ? 

Statement. — Let  x  =  the  amount  of  the  bill ;  then  \x  and  \x 
had  been  paid.  Taking  these  amounts  from  the  bill  we  have 
x—\x—\x.  But  this,  by  the  problem,  was  $92.  Hence  x—\x—\x 
=  93.     .'.  X  =  140. 

7.  The  sum  of  2  numbers  is  180  and  the  difference  10. 
What  are  the  numbers  ? 

Statement. — Let  x  =  the  less  number :  then  a;  + 10  is  the  greater, 
and  a;  +  a; +  10  —  180.  .*.  x  =  85,  and  a;  +  10  =  95,  and  the  two  num- 
bers are  86  and  95. 

8.  The  sum  of  two  numbers  is  5760,  and  their  difference 
is  ^  the  greater.     What  are  they  ? 

Statement. — Let  x  =  the  greater;  then  5760— a;  is  the  less,  and 

jT— (5760— a;)  =  ~.     .'.  x  =  3456,  and  the  less  is  2304. 
3 

9.  A  man  divided  80  cents  among  four  beggars ;  to  the 
first  two  he  gave  equal  amounts,  to  the  third  twice  as  much 
as  to  one  of  these,  and  to  the  fourth  twice  as  much  as  to 
the  third.     How  much  did  he  give  to  each  ? 

The  equation  is  a;+a:  +  2a;  +  4a;  =  80.  Whence  he  gave  the  first 
and  second  10c.  each,  the  third  30c.,  and  the  fourth  40c. 


APPLICATIONS.  241 

10.  A,  B,  C,  and  D,  invest  $4755  in  u  speculation.  B 
furnishes  3  timps  as  much  as  A,  C  as  much  us  A  and  B 
together,  and  D  as  much  as  C  and  B.  How  much  does 
each  invest  ? 

Ani<.,  A,  *317  ;  B,  $951 ;  C,  $1268 ;  and  D,  $2219. 


THE    SAME    PROBLEMS    WITH    THE    LITERAL 
NOTATION. 

11.  What  number  is  that,  to  n  times  which  if  m  be 
added,  the  sum  is  s  ? 

The  equation  is  nx-^m  =  s,  Ans., 

Scholium. — To  make  this  conform  to  Prob.  1,  we  call  «  =  83, 

HO        1          .^           '*—'"      82—18       _„     r»  .^x.                8—^ 
m  =  18,  and  w  =  2.     .'.  =  — - —  =  32.    But  the  answer 

'  71  2  n 

applies  just  as  well  to  any  other  problem  of  the  kind,  no  matter 

what  numbers  are  involved.     Thus,  let  the  problem  be, — What 

number  is  that  to  5  times  which  if  20  be  added,  the  sum  is  100? 

-^o.  ««.        1  ►      «r,  s—m      100—20 

Now  s  =  100,  m  =  20,  and  n  =  5.     Whence  = —  =  16. 

n  5 

We.  therefore^  nee  that is  a  general  answer  to  all  such  problems. 

n 

12.  What  number  is  that,  to  a  times  which  if  b  be  added, 
the  sum  is  c  times  the  number  ? 

b 


Ans.,  Equation,  ax-{-b  =  ex.    .-.  The  number  is 


c—a 


Queries. — How  is  this  adapted  to  Prob.  2  ?    What  other  problems 
can  you  state  to  which  this  value  of  x  affords  an  answer  ? 

13.  What  number  is  that,  a  times  which  exceeds  t  times 

0 

itbyw? 

Equation,  ax  —  t  =  m.    .-.  x  =  -7 — -  is  the  number. 

Queries. —How  is  this  adapted  to  Prob.  3  ?   What  other  problems 


242  SIMPLE    EQUATIONS. 

can  you  state  to  which  -r — r  affords  an  answer  ?    Repeat  the  same 
inquiries  after  each  of  the  seven  following  exanfples. 


14.  The  property  of  two  persons  is  $m,  and  one  owns  n 

times  as  much  as  the  other.     How  much  has  each? 

.  m         ,    7?in 

A71S.,  --—  and  t—-- • 
l-f-/i  1-fn 

15.  What  number  is  that  to  which  if  —th  and  -th  of 

m  n 

itself  be  added,  and  then  a  deducted,  the  result  is  J  ? 

{a^h)mn 


Ans.f 


mn-^in-\-n 


16.  After  paying  — th  and  -th  of  a  bill,  a  remained  du^. 

What  was  the  bill  ? 

cnnn 


Ans., 


mn—n—m 


17.  The  sum  of  two  numbers  is  s,  and  their  difference  d. 

What  are  the  numbers  ? 

.        s  +  d      J  s^d 
Ans.,  -ir-  and  — ;r — 


33 •  Cor. — Observe  that  the  solution  of  this  problem  proves 
the  very  useful  theorem  : 

Half  the  sum  plus  half  the  difference  of  two  quantities  is  the 
greater  of  the  quantities,  and  half  the  sum  minus  half  the  dif- 
ference is  the  less. 

Thus,  the  sum  of  two  numbers  is  20  and  their  difference  13. 
What  are  the  numbers ?  h  +  o  =  10 4-0=16, the  greater;  and  ^^x 
=  10—6  =  4,  the  less, 


APPLICATIONS.  243 

18.  The  sum  of  two  numbers  is  s  and  their  difference  is 

— th  of  the  greater.     What  are  the  numbers. 
m  ° 

_       ..  /        \       ^       4  "^*       s{m—l) 

19.  A  man  divided  ^i^^^  among  4  beggars;  to  the  first  two 
he  gave  equal  amounts,  to  the  third  m  times  as  much  as  to 
each  of  these,  and  to  the  fourth  in  times  as  much  as  to  the 
third.     How  much  did  he  give  to  each  ? 

A  ns.,  To  each  of  the  first  two  $- ^ ,  to  the  third 

am  J  i     ii     J.      ii    A      «wi2 

,,  and  to  the  fourth  1^ 


20.  A,  B,  C,  and  D  invest  $s  in  a  speculation.  B  fur- 
nishes m  times  as  much  as  A;  C  as  much  as  A  and  B 
together,  and  D  as  much  as  C  and  B  together.     How  much 

does  each  invest  ? 

A         .   .       .  ,  s         „       fns       ^    s{l-\-m) 

Ans^  Afurmshes  3^-^;  B,  3^^-^;  C,  -^-^^■, 

3  + 4m 


21.  At  a  certain  election  943  men  voted,  and  the  candi- 
date chosen  had  a  majority  of  65  votes.  How  many  voted 
for  each?  Ans.,  439  and  504. 

22.  A  farmer  has  two  flocks  of  sheep,  each  containing 
the  same  number.  From  one  he  sells  39,  and  from  the 
other  93,  and  then  finds  just  twice  as  many  in  one  flock 
as  in  the  other.  How  many  did  each  flock  originally 
contain?  Ans.,  147. 

23.  A  person  spends  J  of  his  income  in  board  and  lodg- 
ing, -J  in  clothing,  and  ^  in  charity,  and  has  $318  left. 
What  is  his  income  ?  Ans.j  $720. 


344  SIMPLE   EQUATIONS. 

24.  From  a  bin  of  wheat  J  was  taken,  and  20' bushels 
were  added.  After  this  ^  of  what  was  then  in  the  bin  was 
sold,  and  J  as  much  as  then  remained  +  30  bushels  was  put 
in,  when  it  was  found  that  the  bin  contained  just  ^  as  much 
as  at  first.     How  much  did  it  contain  at  first  ? 

Suggestion. — Calling  x  the  amount  in  the  bin  at  first,  after  taking 
out  I,  there  remains  ^x.  To  this  add  20  bushels  and  there  is  in  the 
bin|x  +  30.  After  selling  i  of  this  f  remains,  or  |(|^a;  + 20).  J  of 
this  is  Kf^  +  20).  .-.  |(^a;+20)+Kf'»+30)  +  30  =  ^aj.  Whence 
X  =  560,  the  answer. 

25.  A  person  has  a  hours  at  his  disposal;  how  far  may 

he  ride  iu  a  coach  which  travels  b  miles  per  hour,  and  yet 

have  time  to  return  on  foot  walking  at  c  miles  per  hour  ? 

abc 
Ans.,  J—-' 

26.  After  paying  -th  of  my  money^  and  then  -th  of  what 

remained,  I  had  %a  left.     How  much  had  I  at  first  ? 

Equation,  x — x—{x a:)-  =  a, 

amn 


Ans.f 


27.  A  boy,  being  asked  his  age,  replied,  11  years  are  7 
years  more  than  f  of  my  age.     How  old  was  he  ? 

Statement.— Let  x  =  his  age.     Then  |a;+7  =  11.     .-.  «  =  10. 

28.  A  boy,  being  asked  how  many  sheep  his  father  had, 
replied,  that  40  were  5  less  than  |  of  his  father's  flock.  How 
many  sheep  had  his  father  ?  Arts.,  60. 


29.  If  A  can  perform  a  piece  of  work  in  10  days, 
and  B  in  8  days,  in  what  time  will  they  perform  it 
together  ? 


APPLICATIONS.  Hr) 

Statement.  —Let  x  =  the  number  ot  <lays.  Then  since  A  can  do 
the  work  in  10  days,  in  1  day  he  will  do  3*^,  and  in  x  days  '—  ofit. 

In  like  manner  B  will  do  '-  .     Hence  —  +  -  =  [fiW  the  work)  or  1. 

o  l"        o 

This  1  sometimes  troubles  pupils.  But  let  them  consider  that  if 
two  of  us  do  a  piece  of  w(»rk,  one  doing  §  and  the  other  J,  if  i/ie 
ioorkis  aU  done,  the  sum  of  the  parts  done  is  1. 

30.  There  is  a  certain  piece  of  work  which  A  and  B  can 
do  in  8  days ;  but  A  and  C  can  do  it  in  0  days,  or  B  and  C 
in  10  days.  How  long  would  it  take  any  one  of  them  to 
do  it  alone  ?     How  long  if  all  work  ? 

Suggestion.     «  —  q  is  the  difference  between  what  B  and  C  can 
6       8 

do  in  a  day;  and  --   is  the  sum  of  what  they  can  do  in  a  day. 

(See  33.) 

Ans.,  A  in  lO^-J  days,  C  in  14^,  B  in  344,  and  all  in  5^^. 

31.  A,  B,  and  C  can  do  a  piece  of  work  in  4  days,  which 
A  alone  can  do  in  12,  and  B  alone  in  8  days.  C  begins  the 
work  alone,  and  is  joined  after  2  days  by  A,  and  they  work 
together  2  days  more.  A  and  0  being  then  called  off,  how 
long  will  it  take  B  to  finish  the  work?        Ans.,  5^  days. 

32.  A  performs  f  of  a  piece  of  work  in  4  days  ;  he  then 
receives  the  assistance  of  B,  and  the  two  together  finish  it 
in  G  days.  Required  the  time  in  which  each  could  have 
done  it  alone  ? 

Suggestion. — How  much  does  A  do  in  1  day  ?    Let  x  =  the  time 

B  would  require  to  do  it  all.    The  equation  is  tt  +  -  =  s-  ^  could 

do  it  in  14,  and  B  in  21  days. 

33.  A  vessel  can  be  emptied  by  3  taps  ;  by  the  first  alone 
it  could  he   emptied    in   80   minutes,    by   the   second  in 


M6  SIMPLE    i:QtJATIOiNr. 

200  minutes,  and  by  the  third  in  5  hours.     In  what  time 
will  it  be  emptied  if  all  the  taps  are  opened  at  once  ? 

Ans.,  48  minutes. 

31.  A  can  do  a  piece  of  work  in  a  days  ;  B  in  5  days, 
and  0  the  same  piece  in  c  days.  In  how  many  days  will 
they  finish  it,  when  all  work  together  ? 

abc 


Ans.i 


ab-\-bc-\-  ac 


35.  Tliree  men  A,  B,  and  0  are  employed  on  a  certain 
piece  of  work.  A  and  B  can  do  it  in  a  days;  A  and  C  in 
h  days,  and  B  and  0  in  c  days.  How  long  would  it  take 
each  to  do  it  alone  ?  How  long  would  it  take  them  to  do 
the  work  together  ? 

Ans.,  A  m  7—^ 1  days,  B  in  7 j- , 

hc-\-ac—ab     "^  bc  +  ab—ac 

^  .  2abc  in-       .1       •  ^(^^0 

C  m  -^ r-  J  and  all  tosrether  m  -^ =-  • 

ab-{-ac—bc  ab-\-ac  +  oc 

Suggestion. — Notice  the  singular  symmetry  of  these  answers. 
Such  symmetry  is  common,  and  sometimes  becomes  very  useful  in 
complicated  processes. 


34,  Scholium.-  It  is  not  always  expedient  to  use  x  to  represent 
the  number  sought.  The  solution  is  often  simplified  by  letting  x  he 
taken  for  some  number  from  which  the  one  sought  is  readily  found, 
or  by  letting  2rc,  3x,  or  some  multiple  of  x  stand  for  the  unknown 
quantity.     The  latter  expedient  is  often  used  to  avoid  fractions. 

36.  There  is  a  fish  whose  head  is  9  inches  long ;  the  tail 
is  as  long  as  the  head  and  half  the  body,  and  the  body  is 
as  long  as  the  head  and  tail  together.  What  is  the  length 
of  the  fish?  Ans.,  72  in. 


APPLICATIONS.  H7 

Suggestion. — Let  x  =  the  lenn^tli  of  the  body;  then  9  +  ^a;  =  the 
length  of  the  tail.     The  equation  is  x  =  IS  +  ^a;.    If  a;  =  the  -whoU' 

X  X 

length  the  equation  is9  +  -  +  9+j  =  a'. 

37.  A  general  whose  cavalry  was  J  of  his  infantry,  after 
a  defeat  found  that  before  the  battle  -^^  of  his  infantry  less 
120,  and  -^^  of  his  cavalry  plus  120  had  deserted.  After 
the  battle  he  found  J  of  his  whole  original  army  in  garri- 
son, I  on  the  field,  and  that  the  rest  of  those  engaged  were 
either  taken  i)risoners  or  slain.  Now  300  plus  the  number 
slain  and  captured  was  ^  the  infantry  he  had  at  first.  Of 
how  many  did  his  army  consist  originally  ? 

Suggestion. — Let  x  =  the  number  of  cavalry ;  then  Sec  =  the 
infantry,  and  4:X  =  the  whole  army. 

Ans.y  Whole  army  3600  men ;  viz.,  900  ca.valry,  and  2700 
infantry. 

38.  A  shepherd,  in  time  of  war,  was  plundered  by  a  party 
of  soldiers  who  took  ^  of  his  flock  and  J  of  a  sheep  more ; 
another  party  took  J  of  the  remainder  and  J^  of  a  sheep ; 
and  a  third  party  took  ^  of  the  last  remainder  and  ^  of  a 
sheep,  when  he  had  but  25  sheep  left.  How  many  had  he 
at  first. 

Suggestion.  -Letting  12a;  =  the  number  in  his  flock  at  first,  Qx 
— I  is  what  remained  after  the  first  plundering,  Qx—^  after  the 
second,  and  3.c— |  after  the  third.  The  equation  is  8x— |  =  25. 
.-.  12.r  =  103,  the  number  in  his  flock  at  tirst.  The  pupil  should 
notice  that  he  wants  the  value  of  12x  instead  of  z. 

39.  A  cask,  A,  contains  12  gallons  of  wine  mixed  with 
18  gallons  of  water  ;  another,  B,  contains  9  gallons  of  wine 
mixed  with  3  gallons  of  water.  How  many  gallons  must 
be  drawn  from  each  to  make  a  mixture  of  7  gallons  of  wine 
and  7  of  water? 

Suggestion,  f  of  the  mixture  in  A  is  wine,  and  |  water;  and  in 
B  I  is  wine  and  ^  water.  Let  x  =  the  number  of  gallons  to  be  drawn 
from  A;  then  14— a:  represents  the  number  to  be  drawn  from  R 


248  SIMPLE     EQUATIONS. 

Of  the  first,  |a;  is  wine,  and  fx  is  water;  of  the  second  f(14— ic)  is 
wine,  and  |^(14— a;)  is  water.  But  in  this  new  mixture  the  wine  and 
water  are  equal.     .-.  faJ  +  f(14— «)  =  |«  +  i(14— a-). 

40.  In  the  composition  of  a  quantity  of  gunpowder  the 
nitre  was  10  lbs.  more  than  |  of  the  whole,  the  sulphur 
was  4|  lbs.  less  than  -J-  of  the  whole,  and  the  charcoal  3  lbs. 
less  than  |  of  the  nitre.  What  was  the  amount  of  the  gun- 
powder ?  Ans.y  69  lbs. 

Suggestion,  x  being  the  whole,  the  nitre  is  |aj-f  10,  the  sulphur 
|aj— 4|,  and  the  charcoal  |(|a?+ 10)— 2. 

41.  Several  detachments  of  artillery  divided  a  certain 
number  of  cannon  balls.  The  first  took  72,  and  -J  of  the 
remainder ;  the  next  144  and  J  of  the  remainder ;  the  third 
216,  and  -J  of  .the  remainder ;  the  fourth  288,  and  -J  of 
those  that  were  left ;  and  so  on ;  when  it  was  found  that 
the  balls  had  been  equally  divided.  What  was  the  number 
of  balls  and  detachments  ? 

Ans.,  4608  balls,  and  8  detachments. 

42.  A  gentleman  bequeatlied  his  property  as  follows  :  To 
his  eldest  child  he  left  $1800,  and  }  of  the  rest  of  his 
property  ;  to  the  second,  twice  that  sum  and  |  of  what  then 
remained ;  to  the  third,  three  times  the  same  sum  and  i 
of  the  remainder,  and  so  on  ;  and  by  this  arrangement  his 
property  was  divided  equally  among  his  children.  How 
many  children  were  there,  and  what  was  the  fortune  of 
each  ?  A)is.,  5,  and  $9000  the  fortune  of  each. 

43.  A  wholesale  druggist  has  two  grades  of  attar  of  roses; 
the  one  cost  9  dollars  per  ounce,  the  other  5.  He  wishes  to 
mix  both  grades  together  in  such  quantities  that  he  may 
have  50  ounces,  and  each  ounce,  without  profit  or  loss,  may 
be  sold  for  8  dollars.  How  much  must  he  take  of  each 
grade  to  make  up  this  mixture  ? 

Ans,y  .37|  ounces  of  the  best,  12^  of  the  other. 


APPLICATIONS.  249 

44.  Let  the  price  ol  the  best  grade  in  the  preceding 
problem  =  a  dollars,  the  price  of  tiie  poorest  =  b  dollars, 
the  number  of  ounces  in  the  mixture  =  n,  and  the  price 
of  the  mixture  =  c.  How  many  ounces  of  each  kind  must 
he  use  ? 

.          («— c)n                 -  ^-                 .         ,  {c—b)n     ,   ^, 
Ans.y — -  ounces  of  the  poorest,  and f-  of  the 

other. 

45.  A  father,  who  has  three  children,  bequeaths  his 
property  in  the  following  manner :  To  the  eldest  son 
he  leaves  %1,000,  together  with  the  4th  part,  of  what 
remains ;  to  the  second  he  leaves  12,000,  together  with 
the  4th  part  of  what  remains  after  the  poiidon  of  the 
eldest  and  12,000  have  been  subtracted  from  the  estate; 
to  the  third  he  leaves  |?3,000,  together  with  the  4th  part 
of  what  remains  after  the'  portions  of  the  two  other  sons 
and  $3,000  have  been  subtracted.  The  property  is  found 
to  be  entirely  disposed  of  by  this  arrangement.  What  was 
the  amount  of  the  property  ?  Anf^.,  $9,000. 

46.  A  father,  who  has  three  children,  bequeaths  his 
property  in  the  following  manner :  To  the  eldest  son 
be  leaves  a  sum  a,  t^ogether  with  the  nth.  part  of  what 
remains ;  to  the  second  he  leaves  a  sum  2«,  together  with 
the  n{\\  part  of  what  remains  after  the  portion  of  the 
eldest  and  2«  have  been  subtracted  from  the  estate  ;  to  the 
third  he  leaves  a  sum  3^,  together  with  the  nth  part  of 
what  remains  after  the  portions  of  the  two  other  sons 
and  Za  have  been  subtracted.  The  property  is  found  to  be 
entirely  disposed  of  by  this  arrangement.  What  was  the 
amount  of  the  property  ? 

(67i2_4n  +  l)a 

Arts.,-    (-:rip  • 


;^50  SIMPLE     EQUATIONS. 

47.  A  general  arranging  his  troops  in  the  form  of  a  solid 
square,  finds  he  has  21  men  over,  but  attempting  to  add  1 
man  to  each  side  of  the  square,  finds  he  wants  200  men  to 
fill  up  the  square;  required  the  number  of  men  on  a  side 
at  first,  and  the  whole  number  of  troops. 

Restilt,  110  and  12121. 

Query. — Show  that  this  problem  is  the  same  as  the  following? 
The  difference  between  the  squares  of  two  consecutive  numbers  ib 
221.     What  are  the  numbers  ? 

48.  Gold  is  19|  times  as  heavy  as  water,  and  silver  10^ 
times.  A  mixed  mass  weighs  4160  ounces,  and  displaces 
250  ounces  of  water.  What  proportions  of  gold  and  silver 
does  it  contain  ? 

A71S.,  Gold  3377  ounces ;  silver,  783  ounces. 


THE  TRANSLATION  OF  EQUATIONS  INTO  PRAC- 
TICAL   PROBLEMS. 

S5,  Coy.— Since  the  stalement  of  a  problem  is  expressing 
the  conditions  of  the  problem  in  an  equation,  or  in  equations 
(30),  it  follows,  conversely,  that  an  equation  may  be  considered 
as  the  enunciation  in  algebraic  language  of  the  conditions  of  a 
practical  problem. 

EXAMPLES. 

Ex.  1.  Form  problems  of  which  -  —  -  =  6  is  the  state- 
ment. 

Suggestion. — Any  number  of  problems  may  be  stated  which  will 
meet  the  requirements.  Thus,  "  What  number  is  that  whose  third 
part  exceeds  its  fifth  by  6  ?"  Again ;  'A  man  being  asked  his  age  said, 
'  When  I  was  i  as  old  as  I  now  am,  I  Avas  6  years  older  than  when 
I  was  ^  as  old  as  I  am  now  ; '  what  was  his  age  ?"     Or,  again ;  "  A 


APP1>ICATI0NS.  261 

mau  beiiior  engaged  to  work  a  certain  number  of  days,  continued  in 
the  service  ^  of  the  whole  time,  but  during  this  time  was  sick,  and 
lost  a  number  of  days  equal  to  |  of  the  whole  time,  when  finally  he- 
had  to  break  the  engagement,  having  actually  worked  but  G  days. 
What  was  the  whole  period  engaged  for  ?" 

2.  Form  problems  of  which  a;  +  10a;  =66  is  the  state- 
ment. 

3.  Give  similar  translations  of  each  of  the  following  eqr.:> 

X      X      ^  X       X       ^      ^        X       ^     ^         ,, 

tions:  --[--  +  2  =  2:;  - -f  -  =  1  ;  2x---—  =  7;  x  +  ^x 

-760-r  600  =  2000;  x—^^4^  =  8. 

b 

X  X 

4.  Translate  the  equations  3a; -f- :c  =  16000  ;  ^r-f  ^  +  q 
-10  =  100. 

5.  Enunciate  problems  which  will  give  rise  to  the  follow- 

X         X 

ing  equations  :  x—-^  —  ^=92;  x-\-x-{-2x-[-4.x=SO. 
o        7 

[Note. — The  following  examples  give  rise  to  equations  found 
under  \vi,  26.] 

6.  What  number  is  that  which  gives  the  same  quotient 
when  28  is  added  to  its  square  root,  and  the  sum  divided 
by  its  square  root  +4,  as  when  38  is  added  to  the  same 
root,  and  this  sum  divided  by  the  square  root  of  the  num- 
ber +  6? 

7.  A  man  has  a  certain  number  of  square  rods  of  land 
lying  in  a  square  ;  if  12  rods  be  added,  the  whole  being  kept 
in  the  form  of  a  square,  his  plat  is  increased  by  2  rods  on 
a  side.    How  much  land  has  he  ? 

8.  What  number  is  that  to  which  if  12  be  added  its  square 
root  is  increased  by  2  ? 


252  SIMPLE   EQUATIONS. 

9.  If  4  times  a  certain  uuinber  be  increased  by  1,  the 
square  root  of  this  sum  +  twice  the  square  root  of  the  num- 
ber itself,  divided  by  the  difference  of  the  same  quantities 
is  9.    What  is  the  number  ? 

10.  There  is  a  certain  number  to  which  if  its  square  root 
be  added,  and  the  square  root  of  the  remainder  be  taken, 
and  again  the  square  root  be  subtracted  from  the  number, 
and  the  square  root  of  this  remainder  be  taken,  the  differ- 
ence between  the  results  is  1^  times  the  square  root  of 
the  quotient  of  the  number  divided  by  the  number  4-  its 
square  root.     What  is  the  number  ? 

11.  What  number  is  that  from  which  if  32  be  subtracted, 
the  square  root  of  the  difference  is  equal  to  the  square  root 
of  the  number  —^  the  square  root  of  32  ? 

12.  What  number  is  that  from  which  if  a  be  subtracted, 
the  square  root  of  the  difference  is  equal  to  the  square  root 

of  the  number  —^  the  square  root  of  a  ? 

25a 

Ans,,  — . 

13.  What  number  is  that  whose  square  root  H-2ff,  divided 
by  its  square  root  +&,  equals  its  square  root  -t-4«,  divided 
by  its  square  root  -f-3J  ? 


WITH    TWO    UNKNOWN    QUANTITIES. 

^fel€TS0M    II. 


DEFINITIONS. 

SO,  The  preceding  problems  have  all  been  solved  by  a 
single  equation  containing  only  one  unknown  quantity.  In 
some  of  them  several  quantities  have  been  sought,  it  is  true, 
but  we  have  managed  to  represent  these  quantities  by  the 
use  of  a  single  unknown  quantity,  x.  There  are,  however, 
many  problems  in  which  this  is  not  practicable.  In  such 
problems  there  are  ttvo  or  more  quantities  sought,  and  the 
conditions  are  such  as  to  give  rise  to  two  or  more 
equations. 

Illustration. — To  make  this  latter  statement  clear,  consider  the 
following  problem:  A  says  to  B,  "If  |  of  my  age  were  added  to  | 
of  yours,  the  sum  would  be  19J  years."  But,  says  B  to  A,  "  It"  f  ot 
mine  were  subtracted  from  ^  of  yours,  the  remainder  would  be  18f 
years."  Required  their  ages.  Here  are  two  distinct  quantities 
sought;,  viz.,  A's  age  and  B's  age.  Suppose  we  represent  A's  age 
by  X,  and  B's  by  y.  Now  notice  that  there  are  also  two  sets  of  con- 
ditions. 1st,  the  statement  which  A  makes  to  B ;  and,  2i^d,  the 
statement  which  B  makes  to  A.  According  to  the  1st,  w^e  have  the 
equation  ^x  +  fy  —  19| ;  and  according  to  the  2nd,  |«— |y  =  18f. 

tV7.  Independent  Equations  are  such  as  express 
different  conditions,  and  neither  can  be  reduced  to  the 
other. 

38,  Simultaneous  Equations  are  those  which  express 
different  conditions  of  the  same  problem,  and  consequently 
the  letters  representing  the  unknown  quantities  signify  the 


254  SIMPLE     EQUATIONS. 

same  things  in  each.  Each  equation  of  a  set  of  simultane- 
ous equations  is,  therefore,  satisfied  by  tlie  same  values  of 
the  unknown  quantities. 

Illustration. — Thus  in  the  example  above  the  two  equations  ix 
-f|?/  =  19^  and  Ix—^y  =  18]  are  independent  equations,  since  they 
express  different  conditions^  and  neither  can  be  produced  from  the 
other.  But,  since  these  conditions  are  of  the  same  problem,  so  that 
X  in  the  first  equation  means  the  same  as  x  in  the  second,  and  y  in 
the  first,  the  same  as  y  in  the  second,  they  are  simultaneous  equations. 
It  is  evident  that  the  true  values  of  x  and  y  must  satisfy^  or  ve7'ify, 
both  equations.  If,  however,  we  were  to  write  one  equation  from  one 
problem,  and  one  from  anotlier,  while  they  wotild  be  independent, 
they  would  not  be  simultaneous  ;  x  and  y  would  not  mean  the  same 
things  in  the  first  equation  as  in  the  second.  In  fact,  the  equations 
would  be  so  independent,  that  they  would  have  nothing  to  do  with 
each  other. 

39*  Elimination  is  the  process  of  producing  from 
a  given  set  of  simultaneous  equations  containing  two  or 
more  unknown  quantities,  a  new  set  of  equations  in  which 
one,  at  least,  of  the  unknown  quantities  shall  not  appear. 
The  quantity  thus  disappearing  is  said  to  be  eliminated. 
(The  word  literally  means  putting  out  of  doors.  We  use 
it  as  meaning  causing  to  disappear.) 

40,  There  are  Three  Methods  of  Elimination  in 

most  common  use  :  viz.,  by  Comparison,  by  Svbstltution, 
and  1^  Addition  or  SuUraction.  There  is  also  a  very 
elegant  method  by  Undetermined  Multipliers,  which  is 
worthy  of  more  attention  than  it  generally  receives,  but 
which  will  be  reserved  for  the  advanced  course. 

Scholium. — Any  one  of  these  methods  will  solve  all  problems; 
but  some  problems  are  more  readily  worked  by  one  method  than  by 
another,  while  it  is  often  convenient  to  use  several  of  the  methods  in 
the  same  problem,  especially  when  there  are  more  than  two  un- 
known quantities. 


WITH    TWO    UNKNOWN    QUANTITIES.  255 

ELIMINATION    BY    COMPARISON. 

41,  Prob.  1. — Having  given  two  independent,  simul- 
taneous, simple  equations  between  two  unknown  quantities, 
to  deduce  therefrom  by  comparison  a  new  equation  contain- 
ing only  one  of  the  unicnown  quantities. 

Rule. — /.  Find  expressions  for  the  value  of  the 
same  utiknown  quantity  from  each  equation,  in 
terms  of  the  other  unknown  quantity  and  known 
quantities. 

II.  Place  these  two  values  equal  to  each  other,  and 
the  result  will  he  the  equation  sought. 

Demonstration. — The  first  operations  being  performed  according 
to  the  rules  for  simple  equations  with  one  unknown  quantity,  need 
no  further  demonstration. 

2nd.  Having  formed  expressions  for  the  value  of  the  same  unknown 
quantity  in  both  equations,  since  the  equations  are  simultaneous 
this  unknown  quantity  means  the  same  thing  in  the  two  equations, 
and  hence  the  two  expressions  for  its  value  are  equal,    q.  e.  d. 

Scholium. — The  resulting  equation  can  be  solved  by  the  rules 
already  given. 

EXAMPLES 

OF    INDEPENDENT    SLMULTANEOUS    EQUATIONS. 

Ex.  1.  Given  ^x-\-y  =  34  and  4y  +  a;  =  16;  to  find  x 
and  y  and  verify  the  values. 

Model  Solution. 


Operation.  (1)  4.»4-y  =  34 
=  16 
=  16-4y     .-.  y  = 


4 

(2)  iy+x  =  16  .-.  X  =  16-4^ 

84-y 


4 

ix  +  2  =  34  ..  x  =  S 

Explanation. — Transposing  y  and  dividing  by  4  I  have  from  the 

1st  equation,  x  =  — ~  .     Transposing  4y  in  the  2nd  equation  I 
4 

have  X  =  16— 4y.    Now,  since  these  equations  are  assumed  to  be 

simultaneous,  x  means  the  same  thing  in  both  ;  and  since  things  that 


266  SIMPLE   EQUATIONS. 

34— v 
are  equal  to  the  same  thing  are  equal  to  each  other,  — -^  =  16 

4 

— 4y.  From  this  equation  I  find  y  =  2.  Finally,  since  y  is  found 
to  be  2,  putting  2  lor  y  in  the  first  equation,  I  have  4a;+2  =  34 : 
whence  x  =  8. 

Verification.— Substituting  for  x,  8,  and  for  y,  2,  in  the  1st  equa- 
tion I  have  32  +  2  =  34 ;  and  in  the  2nd,  8  +  8  =  16,  we  see  that 
both  equations  are  satisfied. 

2.  Given  5x-\-4:y  =  58,  and  'dx-\-7y  —  G7,  to  lind  x  and 
y,  eliminating  by  comparison.     Verify  the  results. 

Results,    \     ~  J 
\y  =  7. 

3.  Given  l\x-{-^y  =  100  and  4:x—7y  =  4,  to  find  x  and 
y,  eliminating  by  comparison.     Verify  the  results. 

Results,    <     ~~  / 
ly  =  4:. 

A    c^  1  •         o       -^  +  3        ^       3x—2y        , 

4.  Same   as  above,   given  8 —  =  7 u— ^  and 


Suggestion. 

— Observe  the  effect  of  the  —  sign  before  the  com- 

pound  quantities. 

The  value  of  y  from  the  1st  hy  =  — - — ,   and 

from  the  2nd, 

y  = 

160-30a;      „          5  +  7cc      160-30x        , 
.     Hence  -g—  = ,  and  a;=6, 

and  y  =  5. 

5.  Given 

X 

+  ay  —  h 

dx 

+    y  z=z  c 

6.  Given 

ax 

+    |  =  m 

ex 

-  y  =  n 

7.  Given 

ax 

—  by  :=  m 

ex    -\-  dy  •=  n 

8.  Given  ax-^-hy  —  m,  and  cx-\-dy  =  n,  to  find  x  and  y, 
eliminating  by  comparison. 

Suggestion. — After  having  found  the  value  of  a;  or  y  by  compari- 
son, find  the  value  of  the  other  in  the  same  way. 


WITH     TWO    UNJLNOWK    QUANTITIES.  267 


ELIMINATION    BY    SUBSTITUTION. 

4'i.  Prob.  2.— Having  given  two  independent,  simulta- 
neous, simple  equations,  between  two  unknown  quantities, 
to  deduce  therefrom  by  substitution  a  single  equation  with 
but  one  of  the  unknown  quantities. 

Rule. — /.  Find  frovi  one  of  the  equations  the  value 
of  the  unlcnowrh  quantity  to  he  eliminated,  in  terms  of 
the  other  unknown  quantity  and  knoivn  quantities. 

II.  Substitute  this  value  for  the  same  unknown  quan- 
tity in  the  other  equation. 

Demonstration.- -The  first  process  consists  in  the  solution  of  a 
simple  equation,  and  is  demonstrated  in  Art.  20. 

The  second  process  is  self-evident,  since,  the  equations  being 
simultaneous,  the  lettei*s  mean  the  same  thing  in  both,  and  it  does 
not  destroy  the  equality  of  the  members  to  replace  any  quantity  by 
its  equal,     q.  e  d. 

EXAMPLES. 

Ex.  1.  Given  the  independent,   simultaneous   equations 

* ■^L3'  _  ^iry  =  8,  and  ^  +  ?^i^  =  11,  to  find  x and y, 
2  6  o  4 

eliminating  by  substitution.     Verify  the  results. 

Model  Solution. 

Operation.  (1)    ^  -  fzJ^  =  8 

3x4-3y— 2x  +  2y  =  48 
x-\-5y  =  48 

X  =  48— 6y 

^^^    "a"  "^  "4   ~  ^^ 

(8)    ^8-5y+y  ^  48-5y-y  ^  ^^ 
3  4 

48-4y      U-3y  _ 
8  2       ~ 

96-8y  +  72-9;y  =  66 

-I7y  =  -103 
y  =  6 


258  SIMPLE   EQUATIONS. 

(4)    ^«-^-i^  =  8 

.-.     a;=18 

Explanation. — Taking  equation  (1)  I  clear  it  of  fractions  and 
solve  it  for  cc,  finding  that  x  =  48— 5y.  Now  if  I  take  the  2d 
equation  and  substitute  this  value  for  «,  I  shall  have  a  simple 
equation  with  only  the  unknown  quantity  y  in  it.  This  substitu- 
tion does  not  destroy  the  equality,  since  the  equations  are  simul- 
taneous (3§).      Making    the    substitution    I    have    — ^-7^ — - 

o 

H j^ — -  =  11.      Reducing  this  equation  by  the  method   for 

simple  equations  with  one  unknown  quantity,  I  find  y  =  G. 

Finally,  resuming  (1)  I  substitute  this  value  for  y,  which  evi- 

x-uQ       X 6 

dently  does  not  destroy  the  equation,  and  have  — — -  =  8. 

Solving  (4),  which  is  now  a  simple  equation  with  one  unknown 
quantity,  I  find  x  =  18.  (Instead  of  taking  (1)  in  its  first  form  it 
would  be  better,  because  so  much  shorter,  to  take  its  reduced  form, 
X  =  48— 5y.    .-.  X  =  18.) 

[Note. — The  student  should  keep  a  sharp  lookout  for  oppor- 
tunities to  efl^ect  such  reductions  of  terms  as  are  made  above  in  the 
equations  following  (3)  and  (4).     In  the  latter  the  process  consists 

in  observing  that  — — -  is  -  +  3,  and —  is  —  -  -f-  2,  hence  the 

2  2  o  o 

first  member  becomes  ^  +  ^  "  5  +  ^>  ^^^  transposing  the  3  and  2, 
2  o 

X         X 

we  have  -  —  -  =  3,  all  of  which  can  readily  be  effected  mentally.] 
2       o 

2.  Given  3x—2y  =  1  and  3y—4:X  =  1,  to  solve  as  above. 

Result,  X  —  6  and  y  =  7. 

1  4-  Ax 

Suggestion. — From  the  second,  y  =  — - — ,  hence  the  first 
becomes   3i» —  =  1.     .-,   a;  =  5.    Taking  the  reduced   form 


WITH  TWO   UNKNOWN   QUANTITIES.  259 

of  the  second,  y  =  -  ^— ,  and  putting  5  tor  x,y  =  7.  Verify  both 
equations. 

3.  Given  ^-^+3  =  ^±l,ancl  S-^-^^1  =  |  +  |. 

Eesul/j  a;  =  12  and  y  =  6. 

4.  Given  |  +  |  =  l,aud  |  +  |  =  1.    Verify. 

Result,  X  =  —6  and  y  =  12. 

K    /-,.        a      b  ,  c       <^ 

5.  Given  -  -\ —  =  m.and  — I-  -  =  n. 

X       y  '         X       y 

Suggestion. — If  we  clear  these  equations  of  Tractions  they  will 

become  quite  complex.    But  multiplying  the  first  by  c  and  the 

,  ,  .         CLC      J)c  ^  ac      ad  -r^         ^-^ 

second  by  a,  we  have  —  H —  =  cm,  and  — | =  an.    From  the 

X      y  X       y 

former  —  =  C7n ,  which  substituted  in  the  latter  for  —  pdves 

X  y  X  ^ 

he      ad  ad— be  ctd—lc         ,_^. 

em 4-  —  =  an,  or  =  an  —  cm.     .-.  y  = .       This 

y       y  y  an-cm 

value  of  y  may  now  be  substituted  in  the  first.  To  get  the  value 
of  a?  from  such  an  equation  as  -  =  — -^ — r--,  simply  divide  1  by  each 
member,  i.  c,  take  their  reciprocals. 

6.  Given  — h  -  =  1,  and  -  H —  =  1. 

X       y  X      y 

Result,  X  =  m-{-n,  and  y  =  m-{-n. 

7.  6mn-  +  |=l,  and  3^^  +  1  =  3. 


Result,  X  =  3a,  and  y  =  —25. 


8.  Given  i  -  -  =  5,  and  -  +  i  =  7. 

X      y  X      y 

9.  Given  -^  =  —^ ,  and  x  -\-  iy  =  3. 


260  SIMPLE   EQUATIONS. 

ELIMINATION  BY  ADDITION  OR  SUBTRACTION.- 

43.  Prob.  3.— Having  given  two  independent,  simulta- 
neous, simple  equations  between  two  unknown  quantities, 
to  deduce  therefrom  by  Addition  or  Subtraction  a  single  equa- 
tion with  but  one  unknown  quantity. 

Rul9. — /.  Reduce  the  equations  to  the  forms  ax  +  by 
=  m,  and  cx-fdy  =  n. 

II.  If  the  coefficients  of  the  quantity  to  he  eliminated 
are  not  alike  in  both  equations,  make  them  so  by 
finding  their  L.  C.  M.  and  then  multiplying  the  mem- 
bers of  each  equation  by  this  L.  C.  M.  exclusive  of  the 
factor  ivhich  the  term  to  be  eliminated  already  con- 
tains. 

III.  If  the  sigjis  of  the  terms  containing  the  quantity 
to  be  elUninated  are  alike  in  both  equations,  subtract 
one  equation  from  the  other,  member  by  member.  If 
these  signs  are  unlike,  add  the  equations. 

Demonstration. — The  first  operations  are  performed  according  to 

the  rules  already  given  for  clearing  of  fractions,  transposition,  and 
uniting  terras,  and  hence  do  not  vitiate  the  equation?.  The  object 
of  this  reduction  is  to  make  the  two  subsequent  steps  practicable. 

The  second  step  does  not  vitiate  the  equations,  since  in  the  case 
of  either  equation,  both  its  members  are  multiplied  by  the  same 
number. 

The  3rd  step  eliminates  the  unknown  quantity,  since,'  as  the 
terms  containing  the  quantity  to  be  eliminated  have  the  same 
nurr^erical  value,  if  they  have  the  sarae  sign,  by  suUrading  the 
equations  one  will  destroy  the  other,  and  if  they  have  different 
signs,  by  adding  the  equations  they  will  destroy  each  other.  The 
result  is  a  true  equation,  since,  If  equals  (the  two  members  of  one 
equation)  are  added  to  equals  (the  two  members  of  the  other 
equation),  the  sums  are  equal.  Thus  we  have  a  new  equation  with 
but  one  unknown  quantity.     (^.  e.  d. 


WITH  TWO   UNKNOWN   QUANTITIEa  261 

EXAMPLES. 

Ex.  1.  Given  5x-\-6y  =  28,  and  .r  +  4^  =  14,  to  eliminate 

by  addition  or  subtraction  and  find  the  values  of  x  and  y» 

Verify  the  results. 

Model  Solution. 
Operation.  (1)        5x  +  Qy  =  2S 

(3)  X  +  4y  =  14 

(8)     lOa;  +  12y  =  56 

(4)  dx  +  12y  =  42 

(5)  7x  =  14,  .-.  x  =  2. 

(6)  2  +  4y  =  14 

4y  =  12,  .-.  y  =  3. 

Explanation. — The  equations  being  in  the  required  form  need  no 
reduction. 

To  eHuiinate  y  I  make  its  coefficients  alike  in  both  equations  by 
multiplying  the  members  of  (1)  by  2,  and  of  (2)  by  3,  thus  obtain- 
ing (3)  and  (4).    This  does  not  destroy  the  equations  by  (Ax.  2). 

Then  subtracting  the  members  of  (4)  from  the  corresponding 
members  of  (3)  I  have  (5),  which  is  a  true  equation  since  the  mem- 
bers of  (3)  have  been  increased  equally  (Ax.  2). 

From  (5)  I  have  a-  =  2. 

Finally,  substituting  2  for  a;  in  (2)  I  have  (6),  which  is  a  true 
equation,  since  the  value  of  the  members  of  (2)  is  not  altered  by 
this  substitution  (Ax.  1). 

From  (6)  I  find  by  previous  methods  y  =  3. 

Veriflcation.— Substituting  2  for  x,  and  3  for  y  in  both  of  the 
original  equations,  I  have 

(1)  10  +  18  =  28 

(2)  2  +  12  =  14 

Whence  I  see  that  both  equations  are  satisfied  for  x  =  2,  and 
J)  =3. 

2.  Given  77a;— 12y  =  289,  and  55a; +  27?^  =  491,  to  find 

jc  and  y. 

Model  Solution. 
Operation.     (1)       77a;  -    12y  =  289     (2)   55a;  +    27y  =  491 
693a;  -  lOSy  =  2601       220a;  +  108y  =  1964 
220a;  +  108y  =  1964 
(3)    913a!  =  4565 


Ji62  simplp:  equations, 

(3)     913a;  =  4565 
«  =  5 
(4)    275  4-  27y  =  491 
27y  =  218 

Explanation.— Since  these  equations  are  of  tbe  required  form,  I 
have  only  to  make  the  numerical  values  of  the  terms  to  be  elimi- 
nated alike.  I  will  eliminate  y,  since  its  coefficients  are  smaller 
than  those  of  a;  and  the  process  will  not  involve  as  large  numbers. 
The  L.  C.  M.  of  13  and  27  is  108,  hence  I  multiply  (1)  by  9,  obtain- 
ing G9Sx-108y  =  2601,  and  (2)  by  4,  obtaining  320a;+1082/:=3l964. 
This  process  does  not  vitiate  the  equations,  since.  Equals  multiplied 
by  the  same  number  give  equal  products.  I  now  observe  that  the 
signs  of  lOSy  in  the  two  equations  are  different,  and  consequently 
that  by  adding  the  corresponding  members  of  the  equations  these 
terms  will  destroy  each  other  and  give  an  equation  in  x  only.  Add- 
ing gives  a  true  equation,  since.  Equals  added  to  equals  give  equal 
sums.  I  therefore  have  913a;  =  4565,  or  a;  =  5.  Substituting  this 
result  in  (2),  I  have  275  +  27y  =  491,  whence  y  =  8. 

44:.  Scholium. — It  is  usually  expedient,  in  examples  involving 
two  unknown  quantities,  to  find  the  value  of  the  second  by  substi- 
tution; but  this  is  by  no  means  always  so.  The  pupil  should 
perform  the  examples  in  several  ways,  if  he  can  discern  no  choice 
of  ways  at  first,  and  then  compare  the  methods  with  reference  to 
practicability. 

o    ^.            .  5a;-|-2y      3y— 12-f  Sr?;       ,       15-f  2a;— 4y 
3.  Given  y+      ^  ^  -  ^ — ^-I— -  =  4 -I-^ ^  , 

and ^  -}-  a;  =  2^ —^ ,  to  find  x  and  y. 

Suggestion.— Cleared  of  fractions  and  reduced  to  their  proper 
form  these  equations  become 

6y  -I-      a;  =  34 
45a'  —  31y  =  25 

Whence  a?  =  4,  and  y  =  5.  %' 


WITH  TWO   UNKNOWN   QUANTITIES.  263 

4.  Given  1  H ^^ ^  =  10 j^ ^  and 

=  5a; ^-^ ,  to  find  the  values  of  x  and  y. 

Results,  ic  =  3  and  y  =  7. 

5.  Given  ^^i^  =  ^^  and  8a;~5«/  =  1,  to  find  the 

values  of  a;  and  y.  Results,  x  =  7  and  y  =  l\, 

4«7.  Scholium. — In  practice  it  often  requires  considerable  dis- 
crimination to  determine  which  of  the  methods  of  elimination  to 
employ.  But,  as  any  one  method  will  solve  all  cases,  the  pupil 
need  not  hesitate  too  long  in  attempting  to  select  the  best  one.  If 
he  sees  any  reason  why  one  method  will  be  better  than  another  in 
the  given  case,  he  will  of  course  use  it;  but,  if  no  such  ground  for 
choice  is  apparent,  it  will  often  be  well  to  try  more  than  one 
method,  and  see  if  one  is  any  more  expeditious  than  another. 

EXAMPLES   FOR   GENERAL    PRACTICE. 

Ex.  1.  Given  dx—by  =  13  and  2x-\-7y  =  81,  to  find  x 
and  y.  Results y  a*  =  16  and  y  =  l, 

X  11  X  1J 

2.  Given  -  -f-  j  =  9  and  j  -f-  ^  =  7,  to  find  x  and  y. 

Results,  a"  =  12  and  y  =  20. 

„    ^.        4:X—3y—7       3a;       2y       5       ,  ?/— 1      x      3v 

3.  Given  Jf— =  _  _  ^^  _  and-^+ - -^ 

-  1  =  ^  +  I  +  10'  *^  ^^^  ^  ^^^  ^' 

Results,  a;  =  3  and  y  =  2. 

*• «'- 1  -  -tf  =  1  -  ^'  +  -i^  -^^-^%' 

to  find  X  and  y. 

Suggestion. — The  first  equation  reduces  at  once  to  7x  =  7  +  lly. 
In  this  case  the  pupil  will  see  that  the  three  methods  of  eliminating 
X  are  almost  identical.  Comparing  the  values  of  7.r,  we  have  \2y 
=  7  +  lly:  or  we  may  call  it  suistituting  the  value  of  72"  found  in 


264  SIMPLE   EQUATIONS 

the  second  equation ;  i.  e.,  12y,  for  7x  in  the  first.     Subtracting  the 
second  from  the  first  we  should  have  0  =  7—y. 

6.  Given h  —  =  10  and =  2,  to  find  z  and  y, 

X        y  X       y  ^ 

10      3 
Suggestion. — The  second  becomes  by  dividing  by  2,  — =  1, 

70      21 

and  by  multiplying  by  7, =  7.      Now  adding  this  to  the 

X        y 

85 
first  we  have  —  =  17,  or  a;  =  5  and  hence  y  =  Z. 

6.  Given  -  -\ —  =  m  and ■=.  n.  to  find  x  and  y. 

X       y  X       y  ^ 

^  ^,         ad      Id       ,         ,  he      M      ,  axl  +  bc       , 

Suggestion. 1 =  dm  and =  hi,  •.-. =  dm 

X        y  ^y  X 

■\-hn  and  x  =  -_ — .     Instead  of  substitutinaj  this  value  of  x,  it 

dm  +  on  ^ 

will  be  less  work  to  eliminate  x  from  the  two  given  equations  as  we 

T  T         rm-  ^        ac      le  ^  ac      ad  i        . 

did  y.    Thus  we  have 1 —  —  cm  and =  an,    and    sub- 

X       y  X        y 

ic+ad  bc+ad 

tractmg, =  cm— an,  .-.  y  = 

y  cm— an 

7.  Given  ix-^-^y  =  14  and  ix-{-^y  =  11,  to  find  x  and  y, 
and  verify. 

Suggestion. — Do  not  clear  of  fractions. 

8.  Given  x  —  -^ — n~^  ~ W~^  — 6 

w— 5       lla;  +  152       Sw  +  l     ^    «   ^  i  a         -^ 

—  ^— —  = -r ^ —  ,  to  find  X  and  y,  and  veniy. 

4  12  2      '  ^'  -^ 

-    ^.         -         16  +  60a;       lexy-W^  ^    ^  .  «     .  o 

9.  Given  8a; — i— -- = --^— ,    and    2 +  62^  + 9a; 

Sy — 1  o-\-2y 

27a;2-12^2  +  38     ,„,,,,  .  , 

=  — ; TT^—^ —  ,  to  find  the  values  of  x  and  y. 

Sx—2y-\-l  ^ 

Suggestion. — Multiply  the   1st  equation  by  5  +  2y,  and  reduce 
before  multiplying  by  3y— 1.     Clear  the  2nd  of  fractions.    Whence 
f—UOx  =  187,  and  15a; +  2y  =  86.     .-.  x  =  2,y  =  3. 


WITH   TWO    UNKNOWN    QUANTITIES.  266 

10.  Given    3r  -^  Gv  +  1  =    — ? — r it-^  ,  and 

^  2a:— 4^-h3         ' 

151— IGa:       9a;y— 110     ,    ^    ,  ^,        , 

3x z , —  =  —^ r—  >  to  find  the  values  of  x  and  y. 

4y— 1  3y— 4  ^ 

Result,  X  =z9,  and  ^  =  2. 


11.  Given   16a;  4-  6 v  —  1  = ^ — — ~ ,  and 

^  8a;  —  3y  4-  2 

-^j ~ — ^r-  =  5  — — -,  to  find  the  values  of  x 

2a;  +  2^  +  3  dx-}-2y—l 

and  y.  Result,  x  =  6.  and  y  =  5. 

12.  Given    (x-\-5)  {y  +  H)  =  (^+1)  (y— 9)  +  112,    and 
2a;-f  10  =  3y  +  l,  to  find  the  values  of  x  and  y. 

Result,  X  =  S,  and  y  =  5. 

13.  Given  bcx  =  cy-2b,  and  l^y  +  -i^^  =  ?^  +  c^a;, 
to  find  the  values  of  x  and  y.     Result,  a;  =  7-,  y  = • 

V      2 

Suggestion.— From  the  1st  equation  «  =  | .     Substitutinpr 

0       c 

this  in  the  2nd,  6'y-)--^— — -''  =  —  +  -JL  ^  2e\      Whence,  trans- 

OC  CO 

posing  and  uniting,  y  =  -^ ^  +        , — ^      or    — ^ —  y 

= X  — :r-~  ,  and  y  — .    Substituting  this  value  of  y  in 

Co  c 

the  Ist  equation  and  reducing,  we  find  ^=  j--     These  equations 

can  be  solved  by  a  variety  of  methods,  but  the  pupil  should  con- 
stantly exercise  his  inventive  genius  to  discover  the  most  expedi- 
tious and  elegant  solutions. 

14.  Given  2a;-f-0.4y  =  1.2,   and  3.4^-0.021/ =  0.01,   to 
find  the  values  of  x  and  y.  Result,  x  =  .02,  y  =  2.9. 

12 


•M6  SIMPLE   EQUATIONS. 

Mc    /.•  \x-\-y  =  18.73 

^^•^^^^^    I  0.56^+13.421^  =  763.4 


APPLICATIONS. 

Ex.  1.  There  are  two  numbers,  such,  that  three  times  the 
greater  added  to  one-third  the  less  is  36  ;  and  if  twice  the 
greater  be  subtracted  from  6  times  the  less,  and  the  remain- 
der divided  by  8,  the  quotient  will  be  4.  What  are  the 
numbers? 

Model  Solution. 
Operation.  Let  x  =  the  greater  number, 

and  y  =  the  less  number. 
Then    (1)  3x  +  ^y  =  36 

(2)  «X.-J?  =  4 

(3)  Qy  +  54:X  =  648 

(4)  6y  -    2x  z=    32 

56a;  =  616 

x  =  n 

(5)  6y  -  22  =  32 

y  =  9 

Explanation. — As  there  are  two  unlcnown  qunntities  involved  in 
this  example,  /.  e.,  the  two  numbers  sought,  I  let  x  represent  the 
greater  and  y  the  less.  There  are  also  two  sets  of  conditi(ms  stated 
in  the  problem :  1st,  3  times  the  greater  added  to  ^  the  less  is  36. 
This,  according  to  the  notation,  is  Sx  +  ^y  =  36,  which  is  the  first 
equation.  The  2d  set  of  conditions  is,  that  twice  the  greater  is  to 
he  subtracted  from  6  times  the  less,  which  is  Qy—2x,  and  this  differ- 

Qy 2X 

ence  divided  by  8,  i.  e.,  — - —  .      This   quotient    is  equal    to    4, 
Hence  the  second  equation,  -~ —  =  4. 


WITH    TWO    rNKNOW>r   QUANTITIES.  26f 

2.  Find  two  numbers,  such,  that  if  the  first  be  increased 

by  a,  it  will  be  m  times  the  second  ;  and  if  the  second  be 

increased  by  &,  it  will  be  n  times  the  first  ? 

J,      -.    a-\-bm        ,  b-\-an 
Result y    T,  and 


mn  —  1  mn — 1 

3.  What  two  numbers  are  those,  to  ^  of  the  sum  of  which 
if  I  add  13,  the  result  will  be  17  ;  and  if  from  \  their  differ- 
ence I  subtract  1,  the  remainder  will  be  2  ?     Verify. 

Ans.y  9  and  8. 

[Note. — In  verifying  the  results  in  such  examples  as  these,  no 
attention  s^liould  be  paid  to  the  equations ;  but  the  results  should 
be  tested  directly  by  the  statement.  Thus,  in  this  example,  \  of 
the  sum  of  9  and  3  is  4.  Adding  13  the  result  is,  as  the  example 
requires,  17.  Again  ^  the  difference  of  9  and  3  is  3.  Subtracting  1, 
the  remainder  is  2,  as  required.] 

4.  What  fraction  is  that,  whose  numerator  being  doubled, 
and  denominator  increased  by  7,  the  value  becomes  f ;  but 
the  denominator  being  doubled,  and  the  numerator  increased 
by  2,  the  value  becomes  \  ? 

Ans,,  |. 

Queries  and  Suggestions. — 'How  many  sets  of  conditions  in  this 
problem  ?  What  are  they  ?  How  many  unknown  (required) 
quantities  ?  What  are  they  ?  Tliere  must  always  be  as  many 
of  one  as  of  the  other.  The  unknown  (required)  quantities 
here  are  the   numerator  and  the  denominator  of  the    fraction. 

If   these    are    called    respectively    x  and   y    the    fraction   is    -. 

2x 
Now,  by  the  first  set  of  conditions,  — ;=  —  |,  and,  by  the  second  set, 

x+2  _ 
2y       *' 

5.  What  fraction  is  that  which  becomes  f  when  its  nu- 
merator is  increased  by  6,  and  J  when  its  denominator  is 
diminished  by  2  ?  Ans.,  ^. 


'MjH  simple  equations. 

6.  If  1  be  added  to  the  numerator  of  a  certain  fraction, 
its  value  is  ^ ;  but  if  1  be  added  to  its  denominator,  its  value 
is  }.     What  is  the  fraction  ?    Verify.  Ans,,  ^. 

7.  There  is  a  certain  number,  to  the  sum  of  whose  digits 
if  you  add  7,  the  result  will  be  three  times  the  left  hand 
digit ;  and  if  from  the  number  itself  you  subtract  18,  the 
digits  will  have  changed  places.  What  is  the  number? 
Verify.  A?is.,  53. 

Suggestion.— The  two  numbers  sought  are  the  two  digits.  Hence 
let  y  =  the  units  digit,  and  x  =  the  tens  digit.  The  number  then 
is  10a;+y.  Just  as  when  6  is  the  units  digit  of  a  number  and  5 
the  tens,  the  number  is  10  x  5  +  6.  Of  course  the  number  would  not 
be  represented  by  xy,  for  this  would  indicate  the  product  of  the 
digits.  (See  Part  I.,  30,  Second  Law,  Scholium  1st.)  The  first 
conditions  give  2x—y  =  7,  and  the  second  lOcc  +  y— 18  =  lOy  +  a;, 
i.  e.,  the  units  becomes  the  tens  figure  and  the  tens  becomes  the 
units. 

8.  A  certain  number  of  two  digits  contains  the  sum  of  its 
digits  four  times  and  their  product  twice.  What  is  the 
number?  Ans.,  36. 

9.  There  is  a  number  consisting  of  two  digits  ;  the  num- 
ber is  equal  to  3  times  the  sum  of  its  digits,  but  if  the  num- 
ber be  multiplied  by  3,  the  product  equals  the  square  of  the 
sum  of  its  digits.     What  is  the  number  ?    Verify. 

10.  A  number  consisting  of  2  digits,  when  divided  by  4 
gives  a  certain  quotient  and  a  remainder  of  3  ;  when  divided 
by  9,  gives  another  quotient  and  a  remainder  of  8.  Now, 
the  digit  on  the  left  hand  is  equal  to  the  quotient  which 
was  obtained  when  the  number  was  divided  by  9  ;  and 
the  other  digit  is  equal  to  ^  of  the  quotient  obtained 
when  the  number  was  divided  by  4.  What  is  the  number? 
Verify. 


WITH   TWO    L'NKNOWN   QUANTITIES.  269 

11.  A  farmer  parting  with  his  stock,  sells  to  one  person 
9  horses  and  7  cows  for  300  dollars  ;  and  to  another,  at  the 
same  prices,  6  horses  and  13  cows  for  the  same  sum.  What 
was  the  price  of  each  ? 

A?is.,  *2-t  and  $12. 

12.  A  son  asked  his  father  how  old  he  was.  His  father 
answered  him  thus :  If  you  take  away  5  from  my  years,  and 
divide  the  remainder  by  8,  the  quotient  will  ])e  J  of  your 
age  ;  but  if  you  add  2  to  your  age,  and  multiply  the  whole 
by  3,  and  then  subtract  7  from  the  product,  you  will  have 
the  number  of  the  years  of  my  age.  What  was  the  age  of 
the  father  and  son  ? 

A71S,,  53  and  18. 

13.  A  farmer  purchased  100  acres  of  land  for  $2450;  for 
a  part  of  the  land  he  paid  $20  an  acre,  and  for  the  other  part 
$30  an  acre.  How  many  acres  were  there  in  each  part  ? 
Verify. 

Scholium. — Very  many  such  problems  can  be  solved  equally  well 
by  means  of  one  or  of  two  unknown  quantities. 

14.  At  a  certain  election  946  men  voted  for  two  candidates 
and  the  successful  one  had  a  majority  of  558.  How  many 
votes  were  given  for  each  candidate  ?    Verify. 

15.  A  jockey  has  two  horses  and  two  saddles.  The  sad- 
dles are  worth  15  and  10  dollars,  respectively.  Now  if  the 
better  saddle  be  put  on  the  better  horse,  the  value  of  the 
better  horse  and  saddle  will  be  worth  ^  of  the  other  horse 
and  saddle.  But  if  the  better  saddle  be  put  on  the  poorer 
horse,  and  the  poorer  saddle  on  the  better  horse,  the  value 
of  the  better  horse  and  saddle  will  be  worth  once  and  -^ 
the  value  of  the  other.  Required  the  worth  of  each 
horse  ?  Result,  65  and  50  dollars. 

16.  A  sum  of  money  was  divided  equally  among  a  certain 
number  of  persons  ;  had  there  been  four  more  persons,  each 


270  SIMPLE   EQUATIONS. 

would  have  received  one  dollar  less,  and  had  there  been  four 
fewer,  each  would  have  received  two  dollars  more  than  he 
did :  required  the  number  of  persons,  and  what  each  received? 
Verify. 

Suggestion.    -  =  — T  +  l,and-  =  — -—2.    Rence  xu+4:x.- xy 
««  y      y  +  4:  y      y-4:  ^  ^ 

+2/-  +  4y,   and  xy—4:X  =  xy—2y^-\-Sy,   or  4a;  =  y"^+4y,  and   —2x 

=  —y^+Ay.    Adding,  2x  =  Sy. 

17.  A  farmer  hired  a  laborer  for  ten  days,  and  agreed  to 
pay  him  $12  for  every  day  he  labored,  and  he  was  to  forfeit 
$8  for  every  day  he  was  absent.  He  received  at  the  end  of 
his  time  $40.  How  many  days  did  he  labor,  and  how  many 
days  was  he  absent  ?    Verify. 

18.  A  boatman  can  row  down  stream  a  distance  of  20 
miles,  and  back  again,  in  10  hours,  the  current  being  uni- 
form all  the  time ;  and  he  finds  that  he  can  row  2  miles 
against  the  current  in  the  same  time  that  he  rows  3  miles 
with  it.     Required  the  time  in  going  and  returning. 

Result,  4  and  6  hours. 

Suggestion. — If  aj  and  y  are  tbe  times  of  rowing  down  and  up, 
respectively,  at  what  rate  does  he  row  down  ?  At  what  rate  up  ? 
Twice  one  of  these  rates  equals  3  times  the  other. 

19.  A  and  B  together  could  have  completed  a  piece  of 
work  in  15  days,  but  after  laboring  together  6  days,  A  was 
left  to  finish  it  alone,  which  he  did  in  30  days.  In  how 
many  days  could  each  have  performed  the  work  alone  ? 

Ans.f  50,  and  2 If  days. 

Suggestion. — If  «  represent  the  number  of  days  A  would  require 
to  do  it  alone,  and  y  the  number  B  would  require,  how  much 
would  each  do  in  a  day?  How  much  both?  How  much  would 
they  do  in  6  days  ?  How  much  would  remain  to  be  done  by  A  alone  ? 
How  much  would  A  do  in  30  days  ?  In  resolving  these  equations 
do  not  clear  of  fractions. 

20.  Two  pipes,  the  water  flowing  in  each  uniformly,  filled 


WITH   TWO    UNKNOWN   QUANTITIES.  271 

a  cistern  containing  330  gallons,  the  one  running  during 
5  hours,  and  the  other  during  4 ;  the  same  two  pipes,  the 
first  running  during  two  hours,  and  the  second  three,  filled 
another  cistern  containing  195  gallons.  The  discharge  of 
each  pipe  is  required.    Verify. 

21.  If  I  were  to  enlarge  my  field  by  making  it  5  rods 
longer  and  4  rods  wider,  its  area  would  be  increased  240 
square  rods  ;  but  if  I  were  to  make  its  length  4  rods  less,  and 
its  width  5  rods  less,  its  area  would  be  diminished  210  square 
rods.   Eequired  the  present  length,  width,  and  area.   Verify. 

22.  A  farmer  sells  a  horses,  and  b  cows  for  ^m ;  and  at  the 
same  prices  «i  horses,  and  bi  cows  for  ^//i,;  what  is  the  price 
of  each?    Apply  the  results  to  Ex.  11.     See  (30)  Part  1. 

Ans.,  Of  a  horse  $~^ r  ;  of  a  cow  $ — j -r-- 

aOi—aiO  axb—abx 

[Note. — Observe  the  symmetry  of  such  results.  Thus,  in  these 
numeriitors  the  a  and  5  change  places  and  in  the  denominators  the 
subscripts  change  letters.] 

23.  A  man  bought  s  acres  of  land  for  %m.    For  a  part  he 

paid  %a  per  acre,  and  for  the  rest  $a,  per  acre.     How  many 

acres  in  each  part  ?  Deduce  from  the  getural  answer  obtained 

in  this  case  the  particular  answers  to  Ex.  13. 

.        m—ayS       J  m—as 
Am., and  acres. 

24.  A  waterman  rows  a  given  distance  a  and  back  again 
in  b  hours,  and  finds  that  he  can  row  c  miles  with  the  current 
fore?  miles  against  it:  required  the  times  of  rowing  down  and 
up  the  stream,  also  the  rate  of  the  current  and  the  rate  of 

rowing?        Ans.,  Time  down, ^;  time  up,  — -^ ;  rate 

.  ,  a(c3-^)        .      .        .        flr(c  +  fiO^ 

of  current,  -^^-^  ;  rate  of  rowing,  -^^-^^ 

Peduce  from  these  answers  those  of  Ex.  18, 


Smple'Eiiefims 


"M"^'-^'"^'  """" "" 


^^. 


WITH    MORE   THAN   TWO    UNKNOWN    QUANTITIES. 

SCTI0N    IIL 


4(>,  Prob. — Having  given  several  simple,  simultaneous, 
independent  equations,  involving  as  many  unknown  quanti- 
ties as  there  are  equations,  to  find  the  values  of  the  unknown 
quantities. 

Rule. — /.  Combine  the  equations  two  imd  two  by 
either  of  the  methods  of  elimination,  eliminating  by 
each  combination  the  same  uuJcnown  quantity,  thus 
producing  a  new  set  of  equations,  one  less  in  number, 
and  containing  at  least  one  less  unknown  quantity. 

II.  Combine  this  new  set  two  ajid,  tiuo  in  like  manner, 
eliminating  another  of  the  unknown  quantities. 

III.  Repeat  the  process  until  a  single  equation  is 
found  with  but  one  unknown  quantity. 

IV.  Solve  this  equation  and>  then  substitute  the  value 
of  this  unknown  quantity  in  one  of  the  next  preceding 
set  of  equations,  and  there  will  result  an  equation 
containing  ajvother  single  unknown  quantity,  the 
value  of  which  can  therefore  be  found. 

V.  Substitute  the  two  values  now  found  in  one  of  the 
next  preceding  set,  and  find  the  value  of  the  remain  - 
ing  unknown  quantity  in  this  equation.  Continue 
this  process  till  all  the  unknown  quantities  are  deter- 
mined. 

Scholium  I. — If  any  equation  of  any  set  does  not  contain  the 
quantity  you  are  seeking  to  eliminate  from  the  following  set,  this 


WITH   MORE  THAN  TWO   UN^KNOWN  QUANTITIES.      273 

equation  can  be  written  at  once  in  that  set  and  the  remaining  equa- 
tions combined. 

Scholium  2. — In  eliminating  any  unknown  quantity  from  a  par- 
ticular set  of  equations,  any  one  of  the  equations  may  be  combined 
with  each  of  the  others,  and  the  new  set  thus  formed.  But  some 
other  order  may  be  preferable  as  giving  simpler  results. 

Scholium  3. — It  is  sometimes  better  to  find  the  values  of  all  the 
unknown  quantities  in  the  same  way  as  the  first  is  found,  rather 
than  by  substitution. 

Demonstration  I. — The  combinations  of  the  equations  give  true 
^nations  l)ecause  they  are  all  made  upon  the  methods  of  elimina- 
tion already  demonstrated. 

2.  That  the  number  of  equations  can  always  be  reduced  to  one 
by  this  process,  is  evident,  since,  if  we  have  n  equations  and  com- 
bine any  one  of  them  with  each  of  the  others,  there  will  be  n—\  new 
equations.  Combining  one  of  these  ?i— 1  new  equations  with  all  the 
rest  there  will  result  «— 2.  Hence  n—\  such  combinations  will 
produce  a  single  equation ;  and  as  one  unknown  quantity,  at  least, 
has  disappeared  from  each  set  there  will  be  but  one  left.    q.  e.  d. 


EXAMPLES. 

Sx.  1.  Given  (1,) 

'7x-2z-^^u  =  17, 

(20 
(3i) 

(50 
0  find  the  values  of 

t  +  4.y-2z  =  11, 

5y—dx—2u  =    8, 

-3u-^2t  +  4:yz=    9, 

3z-\-Su  =  33, 

Xy  y,  z,  t,  and  w. 

Operation.       (Ig) 
(2«) 

(4«) 

Model  Solution. 

t^iy-2z=    11 

2<  +  4y-3i^  =      9 

3«  +  8w=    33 

35y_62_5w  =  107 

2nd  set,  from  which 
X  is  absent. 

(1.) 

(23) 
(3a) 

3e  +  8?/=    33]      ^  ,      ^    - 
42,-4.  +  3.=    13  [^  and  «  are  absent. 

1                           (34) 

32  +  8^=      33 
1162-125W  =  -%! 

4th  set,  from  which 
«,  t,  and  y^  are  absent. 

274 


SIMPLE    EQUATIONS. 


(Ig)  1303^  =  3909     .:  u=S 

(IJ  3s  +  34=33  .-.  2  =  3 

(33)  42/-12  +  9  =  13  .'.  y  =  4 

(Ig)       ^  +  16-6  =  11  .-.    t  =  l 

(3i)  20-3a!-6=    8  ::  x  =  2 

Explanation. — I  notice  that  I  have  5  equations  with  5  unknown 
quantities.  From  these  I  wish  to  produce  a  new  set  of  4  equations 
from  which  one  at  least  of  the  unknown  quantities  shall  be  eliminated. 
I  observe  that  x  does  not  appear  in  (2^),  (4^),  and  (5  J,  hence  I  write 
these  as  three  of  the  2d  set  of  equations.  Then  eliminating  x 
between  (Ij)  and  (SJ  I  have  (4g),  and  thus  obtain  the  2nd  set  of  4 
equations  containing  only  4  unknown  quantities. 

Again,  as  t  is  contained  in  a  less  number  of  this  set  of  equations 
than  any  one  of  the  other  unknown  quantities,  I  eliminate  it  next ; 
i.  e.,  I  produce  a  3rd  set  which  does  not  contain  it.  As  (^2)  and 
(42)  do  not  contain  t,  I  transfer  them  at  once  to  the  3rd  set ;  and 
then  eliminating  t  between  (I3)  and  (22)  this  set  is  complete,  having 
3  equations  with  3  unknown  quantities. 

Now  eliminating  y  from  this  set  by  combining  (23)  and  (33),  and 
transferring  (I3),  I  have  the  4th  set  of  two  equations  with  only  2 
unknown  quantities.  Combining  these  two  so  as  to  eliminate  z  I 
find  u  =  S. 

Finally,  substituting  3  for  u  in  (I4),  I  find  z  =  d. 
Substituting  3  for  u  and  3  for  z  in  (33),  I  find  y  =  4. 
Substituting  4  for  y  and  3  for  z  in  (Ig),  I  find  t  =  i. 
Substituting  the  values  of  g^  and  ^^  in  (3i),I  find  x  =  2. 


2.  Given  j  x-\-z  =  26 
?/+  2;  =  15 


3.  Given- 


4.  Given 


Sx^4:y  =  24—  2, 

6x-\-  y  =    2;  +  84, 

a;  +  80  =  3«/  +  4;2. 

3w+  a;4-2y—  0  =  22 
^x—  y  +  Sz  =  35 

4:U-^dx—2y  =  19 
2^  +  42^4-22;  =  46 


Values,  - 


Values,  ■ 


Values, 


x=:20, 
y  =  10, 

z  =    6. 

\x  =  n, 

y=z20, 

z  =    S. 


'  U  =  4:, 

X  =  6, 

z  =7. 


WITH  MORE  THAN  TWO   UNKNOWN  QUANTITIES.      275 


5.  Given 


2^3^4 


124, 


Results,  • 


x=    48, 

y  =  120, 

z  =  240. 


Suggestion. — This  may  be  solved  in  the  ordinary  way  by  clearing 
of  fractions,  etc.,  but  the  following  is  far  more  elegant: 

Dividing  the  Ist  by  3  and  the  2nd  by  2  and  subtracting,  we  have 


Subtracting  ^  the  Ist  from  the  3rd 


y 

z 

17] 

+ 

72 

60 

'6 

y 

+ 

2 

=  14 

80 

24 

2nd  set. 


Or, 


j_  g     _17 

360  "^  5  •  60  ~  15 

y         ^     _  "^ 

360'^2n2~6 


Subtracting  — — 


2        6  1 


6.  Given 


n    1 

y    z 


=  1,  and  z  =  240. 
2 


Values,  ■ 


y 


a  +  h—c* 

2 
a—h-J^c' 

2 


2;  = 


^H-c— « 


Suggestion.— Do  not  clear  of  fractions.  Having  found  the  value 
of  one  unknown  quantity,  do  not  get  the  others  by  substitution,  but 
return  to  the  original  equations  and  get  each  in  the  same  manner. 


7.  Given 


r2  3__4_  J_ 
x"^  y  z~12' 
3_4  5_19 
X     y^  z"  24' 

x^  y^  z      2 


Values, 


X  =    6, 

%  =    8. 


276 


SIMPLE    EQUATIONS. 


x-\-a=    y  +  z, 
8.  Given  \  y-\-a  =  2xi-2z, 
z-{-a  =:  3x-\-Sy, 

r    9x-^6y-{-4:Z-\-3v-^2u' 

16a;+4v-|-l  =  0, 

25x—5y-\-z  -\-  6v  —  u 
+  1  =  0, 
x-}-2y-\-4:Z—v  —  2u 
+  1  =  0, 
^x—2y-\-z  —  2v  -\-  u 
+  1  =  0. 


Values, 


z  = 


a 

il' 

Tl' 

7a 
11* 


9.  Given 


■  Values,  ■ 


X  = 


u  = 


10.  Giv'^n 


11.  Given- 


u-{-v-{-x-\-y  =  10, 
u-^v-{-x-\-z  =  11, 
w  +  v  +  ?/  +  2;  =  12, 
w  +  ^  +  ?/  +  ^  =  13, 
y  ■^x-\-y-\-z  =  14. 

3j--l_65     X 

■    4      -  5      2"^  ^' 


Fa^i^e^?,  - 


6X          4:Z 

T"^  3" 

,  5 
=  2^  +  6' 

3:^  +  1 

7 

14^6 

2z 
'21 

+  ^ 
^3 

Values, 


12.  Given 


11a;— lOi/  = 


12y—llz 


x  +  z—2y  _  2;— ?/— 1 
3         ""        2       ' 

3^-        =?/+;2;  +  7. 


Values,  ■ 


169 
"924' 
220 
924' 
89 
924' 
445 
924' 
113 
924* 

=  3, 

=  4, 
=  5, 
=  2, 
—  1, 

=  2, 
=  3, 
=  1. 


r?;  =  10, 


L2! 


11, 

12. 


WITH   MORE  THAN  TWO    UNKNOWN   QUANTITIES.      277 


13.  Given 


10  15 

9x-{-5y—2z     2x-{-y—dz 
12  4 

_7y±z±S 
-       11 "     "^^^ 

5y-\-3z     2x-{-3y—z 
4  12 


-\-2z 


:y-l  + 


3a;  +  2y  +  7 


Values,  ■ 


a;  =  9, 


2/  =  7, 


2=3. 


14.  Given 


4(a;  +  2;)  =  9-2/, 
i{x-z)  =  2y-7. 


Values, 


x  =  7, 
z=3. 


APPLICATIONS. 

Ex.  1.  The  sum  of  three  numbers  is  9.  The  sum  of  the 
first,  twice  the  second,  and  three  times  the  third  is  22.  The 
sum  of  the  first,  four  times  the  second,  and  nine  times  the 
third  is  58.    What  are  the  numbers?      Aiis.,  1,  3,  and  5. 

Suggestion. — How  many  unknown  quantities?  How  many  sets 
of  conditions  ?  What  are  they?  Express  the  first  in  an  equation, — 
the  second, — the  third, 

2.  Five  persons,  A,  B,  C,  D,  and  E  played  at  cards;  after 
A  had  won  half  of  B's  money,  B  one-third  of  C's,  C  one- 
fourth  of  D's,  and  D  one-sixth  of  E's,  they  each  had  $7.50. 
How  much  had  each  to  begin  with  ? 

Ans.,  A,  $2.75 ;  B,  $9.50 ;  0,  $8.25  ;  D,  $8 ;  and  E,  $9. 

3.  There  are  4  men.  A,  B,  C,  and  D,  the  value  of  whose 
estates  is  $14,000 ;  twice  A's,  three  times  B's,  half  of  C's, 


278  SIMPLE    EQUATIONS. 

and  one-fifth  of  D's,  is  $16,000 ;  A's,  twice  B's,  twice  C's, 
and  two-fifths  of  D's,  is  $18,000 ;  and  half  of  A's,  with  one- 
third  of   B's,  one-fourth  of  C's,  and   one-fifth   of   D's,  is 
$4,000.     Kequired  the  property  of  each. 
Ans.,  A's,  $2,000;  B's,  $3,000;  C's,  $4,000;  D's,  $5,000. 

4.  A  number  is  represented  by  three  figures  ;  the  sum  of 
these  is  11 ;  the  figure  in  the  place  of  units  is  double  that  in 
the  place  of  hundreds,  and  when  297  is  added  to  this  num- 
ber, the  sum  obtained  is  represented  by  the  figures  of  this 
number  reversed.     What  is  the  number  ?  Ans.,  326. 

Suggestion. — Letting  x  represent  the  hundreds  figure,  y  the  tens, 
and  s  the  units,  the  number  is  represented  by  lOOx+lOy-^z.  The 
number  with  the  digits  reversed  is  lOOz+tOy-^x. 

5.  A  man  worked  for  a  person  ten  days,  having  his  wife 
with  him  8  days,  and  his  son  6  days,  and  he  received  $10.30 
as  compensation  for  all  three ;  at  another  time  he  wrought 
12  days,  his  wife  10  days,  and  son  4  days,  and  he  received 
$13.20  ;  at  another  time  he  wi'ought  15  days,  his  wife  10 
days,  and  his  son  12  days,  at  the  same  rates  as  before,  and 
he  received  $13.84.     What  were  the  daily  wages  of  each  ? 

Ans.y  The  husband  75  cts. ;  wife,  50  cts.  The  son,  20 
cts.  expense  per  day. 

Suggestion.— The  value  of  the  quantity  representing  the  son's 
wages  is  found  to  be  negative.  Therefore  the  son  produced  the 
opposite  effect  from  wages;  i.  e.,he  was  an  expense. 

6.  Three  masons.  A,  B,  C,  are  to  build  a  wall.  A  and  B, 
jointly,  can  build  the  wall  in  12  days ;  B  and  C  can  accom- 
plish it  in  20  days,  and  A  and  C  in  15  days.  How  many 
days  would  each  require  to  build  the  wall,  and  in  what  time 
will  they  finish  it,  if  all  three  work  together  ? 

A71S.,  A  requires  20  days;  B,  30  ;  and  C,  60  ;  and  all 
three  require  10  days. 


WITH   MORE  THAK   TWO   UNKNOWN   QUANTITIES.      279 

7.  Three  laborers  are  employed  on  a  certain  work.  A 
and  B,  jointly,  can  complete  the  work  in  a  days ;  A  and  0 
require  h  days,  B  and  C  require  c  days.  What  time  does 
each  one,  working  alone,  require  to  accomplish  the  work, 
on  the  condition  that  each  one,  under  all  circumstances, 
does  the  same  quantity  of  work  ?  And  in  what  time  would 
they  finish  it,  if  they  all  three  worked  together  ? 

Am.,  A  reqmres  j^^^^z^  days,  B  ^^^„^_^  days, 

days. 

Deduce  from  these  results  those  of  the  preceding 
example. 

8.  If  A  and  B  together  can  perform  a  piece  of  work  in 
8  days,  A  and  C  together  in  9  days,  and  B  and  C  together 
in  10  days,  in  how  many  days  can  each  alone  perform  the 
same  work? 

Ans.^  A  in  14|f  days,  B  in  17|f  days,  and  C  in  23-^ 
days. 

9.  A  gentleman  left  a  sum  of  money  to  be  divided  among 
his  four  sons,  so  that  the  share  of  the  oldest  was  ^  of  the 
sum  of  the  shares  of  the  other  three,  the  share  of  the 
second  \  of  the  sum  of  the  other  three,  and  the  share  of  the 
third  J  of  the  sum  of  the  other  three  ;  and  it  was  found 
that  the  share  of  the  oldest  exceeded  that  of  the  youngest 
by  $14.  What  was  the  whole  sum,  and  what  was  the  share 
of  each  person  ? 

Ans.,  Whole  sum,  $120 ;  oldest  son's  share,  $40  ;  second 
son's,  $30  ;  third  son's,  $24 ;  youngest  son's,  126. 


280 


SIMPLE    EQUATIONS. 


CO 

o 


Q. 
CO 


DEFINITIONS. 


SYNOPSIS. 

f  Algebra. 
Equation.     Members. 
Independent  Equations.    Simultaneous. 
Transposition.     Elimination. 
Statement.    Solution. 


With  one  unknown  quMitity. 

With  more  than  one  unknown  quantity. 


KINDS   OF    ^^""f\. 
EQUATIONS.  )  Q"adratrc 

^  Higher. 
AXIOiVIS. 

I.   Clearing  of  fractions.    Rm-is.   Dem.   m. 
TRANSFOR-  J  2.  Transposition,     rule.   Detn.   ill. 
MATIONS.    1  3.  Uniting  terms. 

4.  Dividing  by  coefficient  of  unknown  quan. 


Prob.  I.  To  solve  equations 

RuLB.    Dem. 

Prob.  2.  To  free  an  equation  of  radicals 


RuLB.    Dem.  [  Practical  suggestions. 


Rule.    Dem. 


Practical  suggestions. 


LJ 


f  Number  of  methods.    Reason  for  several. 
Prob.  I.   By  comparison,     rule.   Dem. 
Prob.  2.  By  substitution,    rule.   Dem. 
Prob.  3.  By  addition  and  subtraction,    rule.    Dem. 
Prob.  4.  With  several  unknown  quantities.  I  „  ^  , 

RULE.    Dem.  [sch.l,2,a 


Test  Questions. — Upon  what  principle  is  an  equation  cleared  of 
fractions  ?  How  is  it  done  ?  Upon  what  principle  is  elimination 
by  addition  and  subtraction  performed  ?  What  comparison  ?  Sub- 
stitution? Give  the  seven  Practical  Suggestions  upon  solving 
Simple  Equations.  The  six  upon  freeing  of  Radicals.  Give  the 
reason  for  changing  the  signs  of  the  terms  of  a  fraction  having  a 
polynomial  numerator,  preceded  l)y  a  minus  sign,  when  clearing  of 
fractions.  What  is  the  general  method  of  precedure  in  stating  a 
problem  ?  Does  the  statement  involve  a  knowledge  of  anything  but 
algebra  ?  Illustrate.  Upon  what  principle  may  all  the  signs  of  an 
equation  be  changed  ?  (This  may  be  explained  in  at  least  four  dif- 
ferent ways.)  Having  given  the  sum  and  difference  of  two  quanti- 
ties, how  are  the  quantities  found  ?  Prove  it. 


^OPOF^TION 
F\OGF^ESSION 

RATI  0. 

47.  Ratio  is  the  relative  magnitude  of  one  quantity  as 
compared  with  another  of  the  same  kind,  and  is  expressed 
by  the  quotient  arising  from  dividing  the  first  by  the  second. 
The  first  quantity  named  is  called  the  Antecedent,  and  the 
second  the  Consequent,  Taken  together  they  are  called  the 
Terms  of  the  ratio,  or  a  Couplet, 

48,  The  Sign  of  ratio  is  the  colon,  : ,  the  common  sign 
of  division,  -^,  or  the  fractional  form  of  indicating  division. 

Illustration. — The  ratio  of  8  to  4  is  expressed  8 :  4, 8^4,  or  j »  a°y 

one  of  which  may  be  read  "  8  is  to  4,"  or  "  ratio  of  8  to  4."  The 
antecedent  is  8,  and  the  consequent  4.  The  sign  :  is  an  exact 
equivalent  for  -j-,  and  by  many  writers,  especially  the  Germans,  the 
former  is  used  exclusively.  The  sign  :  is,  probably,  a  mere  modi- 
fication of-!-, made  by  dropping  the  horizontal  line,  as  unnecessary. 
Possibly  the  sign  -i-  finds  its  analogy  in  the  fractional  form  for 
expressing  division,  by  considering  the  upper  dot  as  symbolizing  a 
dividend,  and  the  lower  a  divisor. 

40.  Cor. — A  ratio  being  merely  a  fraction,  or  an  unexe- 
cuted problem  in   Division,   of  which  the  antecedent  is  the 


282  RATIO,   PROPORTIOi?^,   AND   PROGRESSION. 

numerator,  or  dividend,  and  the  consequent  the  denominator,  or 
divisor,  any  changes  made  upon  the  terms  of  a  ratio  produce 
the  same  effect  upon  its  value,  as  the  like  changes  do  upon  the 
value  of  a  fraction,  when  made  upon  its  corresponding  terms. 
The  principal  of  these  are, 

1st.  If  both  terms  are  multiplied,  or  both  divided  by  the  same 
number,  the  value  of  the  ratio  is  ISOT  changed. 

Illustration.  16 :  8       =2 

4x16:4x8  =  2 

2nd.  A  ratio  is  multiplied  by  multiplying  the  antecedent 
or  by  dividing  the  consequent. 

8  =  4 
8  =  8 


Illustration.  32 

2x32 

32 


=  8 


3rd.  A  ratio  is  divided  hy  dividing  the  antecedent  or  by 
multiplying  the  consequent. 
Illustration. 


24 

:6 

=  4 

¥ 

:6 

=  2 

24 

2x6 

=  2 

50.  A  Direct  Ratio  is  the  quotient  of  the  antecedent 
divided  by  the  consequent,  as  explained  above.     (47.) 

An  Indirect  or  Reciprocal  Ratio  is  the  quotient  of 
the  consequent  divided  by  the  antecedent,  i.  e.,  the  recipro- 
cal of  the  direct  ratio. 

Thus,  the  direct  ratio  of  6  to  3  is  2,  but  the  inverse  ratio  is  |  or  ^. 
When  the  word  ratio  is  used  without  qualification  it  means  direct 
ratio.  The  inverse  or  reciprocal^  it  will  be  seen,  is  the  ratio  of  the 
reciprocals.  Thus  the  inverse  ratio  of  8  to  4  is  the  ratio  of  |^  to  J, 
or  f 

51.  A  ratio  is  always  writtek  as  a  direct  ratio. 

Thus,  the  inverse  ratio  of  a^  to  &  is  & :  a,  or  -  :  ^ ,  the  latter  being 
expressed  as  the  direct  ratio  of  the  reciprocals. 


RATIO.  283 

52.  A  ratio  of  Greater  Inequality  is  a  ratio  which  is 
greater  tlian  unity,  as  4  :  3.  A  ratio  of  Less  Inequality 
is  a  ratio  which  is  less  than  unity,  as  3 : 4. 

53»  A  Compound  Ratio  is  the  produpt  of  the  cor- 
responding terms  of  several  simple  ratios. 

Thus,  the  compound  ratio  a :  6,  c :  ^,  w :  w,  is  acm :  Mn.  This  term 
corresponds  to  compound  fraction.  A  compound  ratio  is  the  same 
in  eflfect  as  a  compound  fraction. 

54,  A  Duplicate  Ratio  is  the  ratio  of  the  squares,  a 
triplicate,  of  the  cuhes,  a  subduplicate,  of  the  square 
roots,  and  a  subtriplicate,  of  the  ciche  roots  of  two  num- 
bers.   Thus,  «2 :  ]^^  a^:i)S^  ^a  :  ^b,  and  \^a  :  ^/b. 

EXAMPLES. 

Ex.  1.  What  is  the  ratio  of  3am^  to  6am^  ? 

Model  Solution. — Since  the  ratio  of  two  quantities  is  the  quotient 

of  the  antecedent  divided  by  the  consequent,  the  ratio  3am^ :  Qam^  is 

3am»         1 
- — - ,  or  -m. 

2.  What  is  the  inverse  ratio  of  a— J  to  a^—l^  ? 

A71S.,  a  +  b, 

3.  What  is  the  ratio  of  I  to  ^?    Of|:|?    Of  ^^^=^  :  a 

3      2  6    3  a^-^a^ 

_^v  Of  — to?^?  Of  ""'-y' .    ^-^ 

3%        3by  T^+f  x^-xy+y^ 

Answers  to  threes  1,  IJ,  and  1-J-. 

4.  What  is  the  triplicate  ratio  of  6  to  2,  Arts.,  27. 

5.  What  is  the  subduplicate  ratio  of  64  to  16  ?  Ans.,  2. 

6.  What  is  the  compound  ratio  of  3  to  4,  8  to  9,  2  to  6, 
and  4  to  2  ?  Ans.,  4  : 9,  or  ^. 

7.  Reduce  360  :  315  to  its  lowest  terms. 

Suggestion. — This  is  the  same  as  reducing  a  fraction  to  its  lowest 
teniis.     The  result  is  8 :  7. 


284  RATIO,   PROPORTION,   AND   PROGRESSION. 

8.  Reduce  a^  -\-  2a^x  :  a^  to  its  lowest  terms. 

Result,  a-|-2a; :  1. 

9.  Which  is  the  greater,  16  :  15,  or  17  to  14? 

Suggestion. — To  compare  two  fractions,  reduce  them  to  a  common 
denominator.  On  the  same  principle  these  ratios  become  224  :  210, 
and  255 :  310.  Otherwise  perform  the  division  and  compare  the 
quotients.    By  this  method  we  have  1.06 +  ,  and  1.21 +  . 

10.  Which  is  greater,    a^—W\a—l,    or    a^-i-2ab  +  i^'. 

11.  Which  is  least  of  the  ratios  20:17,  22:18,  and 
25  :  23  ? 

12.  Which  is  greater,  «  +  2  :  J^-f  4,  or  «  +  4  :  Ja  +  S  ? 

Ans.f  a  +  4:  :  ^a-\-6ya-{-2  :  ia-\-4:. 

13.  What  is  the  compound  ratio  of  15  :  12,  6  :  7,  and 
9:4?  Ans.,  135  :  56. 

14.  Compound  the  ratios  a^—x^ia^,  a  +  x:i,  and 
h  \  a—x.  Result,  The  duplicate  ratio  of  a-\-xto  a. 


15.  Show  that  the  compound  ratio  of  x-\-y  \  a,  x—y:h, 
and  h  : —  is  1. 


16.  Is  the  compound  ratio  of  3a  +  2  :  6«4-l,  and  2rt  +  3  : 
a +2,  a  ratio  of  greater  or  of  less  inequality,  if  a  is  — f  ?  If 
a  is  2  ?     If  «  is  —  2  ?         ^^s.,— Zero,  Greater,  Infinity. 

17.  Compound  the  following  :  7  :  5,  the  duplicate  of 
4  :  9,  and  the  triplicate  of  3  :  2.  Result,  14  :  15. 

7       ^.|       $'$'$      ,.    -,. 
Suggestion.     5  X  ^^  X  ^^^  =  14  .  15. 

3 

18.  Compound  the  sub-duplicate  of  x^ :  y^,  and  the  tripli- 
cate of  ^/x  :  \^y.  Result,  x^ :  y^. 


PROPORTION.  285 

19.  Compound  the  inverse  ratio  of  ^x-\-*Jy  to  ar— -y, 
and  the  direct  ratio  x-\-'l^xy-^y  :  Vx-{-Vy- 

Result,  x—y, 

20.  Which  is  the  greater,  the  inverse  subtriplicate  ratio 
of  8  to  64,  or  the  direct  duplicate  ratio  of  2  to  3  ? 


Pf^opof^tion. 
^^csf^^^TMM   IL 


55,  Proportion  is  an  equality  of  ratios,  the  terms  of 
the  ratios  being  expressed.  The  equality  is  indicated  by 
the  ordinary  sign  of  equality,  =,  or  by  the  double  colon  : : . 

Thus,  8  :  4  =  6  :  3,  or  8  :  4  : :  6  :  3,  or  8-f-4  =  6h-3,    or   -  =  - 

all  mean  precisely  the  same  thing.     A  proportion  is  usually  read 
thus :  "  as  8  is  to  4  so  is  6  to  3." 

Scholium.— The  pupil  should  practice  writing  a  proportion  in  the 

form  5-  =  3 ,  still  reading  it  "  a  is  to  5  as  c  is  to  dy    One  form 
0      d 

should  be  as  familiar  as  the  other.     He  must  accustom  himself  to 

a       c 
the  thought  that  a:b  ::  c:d  means  t  =  -^  and  nothing  more.   It  will 

be  seen  that  the  language  "  8  is  to  4  as  6  is  to  3,"  means  simply  that 

ft      (\ 

-  r=  - ,  fbr  it  is  an  abbreviated  form  for  saying  that  "  the  relation 

4       8 

which  8  bears  to  4  is  the  same  as  (is  equal  to)  that  which  6  bears 
to  3 ; "  that  is,  8  is  as  many  times  4  as  6  is  times  3,  or  -  =  -  • 

50.  The  Extremes  (outside  terms)  of  a  proportion  are 
the  first  and  fourth  terms.  Tlie  Means  (middle  terms)  are 
the  second  and  third  terms.  Thus,  in  a  :  b  =  c  :  d,  a  and 
d  are  the  extremes,  and  b  and  c  are  the  means. 


286  RATIO,   PROPORTION,   AND   PROGRESSION. 

57.  A  Mean  Proportional  between  two  quantities  is  a 
quantity  to  which  either  of  the  two  bears  the  same  ratio 
that  the  mean  does  to  the  other. 

Thus,  if  7W  is  a  mean  proportional  between  a  and  5,  a  bears  the 
same  ratio  to  m  that  m  does  to  5 ;  i.  e.^  a-.m  ::  m:  K 

58.  A  Third  Proportional  to  two  quantities  is  such  a 
quantity  that  the  first  is  to  the  second  as  the  second  is  to 
this  third  (proportional). 

Thus,  in  the  last  proportion,  5  is  a  third  proportional  to  a  and  m. 
So,  also,  a  is  a  third  proportional  to  5  and  m. 

Scholium. — Notice  carefully  the  language  used  in  the  last  two 
definitions.  We  do  not  say  "a  mean  proportional  ^,"  but  "a 
mean  proportional  between,''^  two  others.  So,  again,  we  say  "  a  third 
proportional  to  two  others."  Moreover,  it  is  necessary  that  the  two 
others  be  taken  in  the  order  named  in  the  statement.  Thus,  if  y  is 
a  third  proportional  to  m  and  n,  m:n  \:n:y.  But,  if  y  is  a  third 
proportional  to  n  and  m^  n-.m  ::  m:y. 

59.  A  proportion  is  taken  by  Inversion  when  the  terms 
of  each  ratio  are  written  in  inverse  order.  Thus,  if 
a  :  b  : :  c  :  d,  hy  inversion  we  have  h  :  a  :  :  d  :  e.  It  is  to  be 
observed  that  in  inversion  the  means  are  made  extremes, 
and  the  extremes  means. 

60.  A  proportion  is  taken  by  Alternation  when  the 
means  are  made  to  change  places,  or  the  extremes.  Thus, 
a  :  b  ::  c  :  d  becomes  by  alternation  either  a  :  c  ::  h  :  d,  or 
d  :  b  ::  c  :  a.  The  appositeness  of  the  term  alternation 
(taking  every  other  one)  is  seen  from  the  fact  that  the  new 
order  is  obtained  by  taking  the  terms  alternately ;  that  is, 
1st  and  3rd,  2d  and  4th  ;  or  4th  and  2nd,  3rd  and  1st. 

61.  A  proportion  is  taken  by  Composition  when 
the  sum  of  the  terms  of  each  ratio  is  compared  with  either 
term  of  that  ratio,  the  same  order  being  observed  in  both 
ratios ;  or  when  the  sum  of  the  antecedents  and  the  sum  of 


PROPORTION.  287 

the  consequents  are  compared  with  either  antecedent  and 
its  consequent.  Thus,  if  a  :  h  :  :  c  :  dy  by  composition 
we  have  a-\-b  :  a  :  :  c-\-d  :  c,  or  a-\-b  :  b  : :  c-\-d  :  d,  or 
a-^c  :  b-^d  ::  a  :  b,  or  a  +  c  :  b-\-d  ::  c  :  d. 

02.  If  the  difference  instead  of  the  su7ri  be  taken  in  the 
last  definition,  the  proportion  is  taken  by  Division. 

03.  Four  quantities  are  Inversely  or  Reciprocally  Propor- 
tional when  the  1st  is  to  the  2nd  as  the  4th  is  to  the  3rd, 
or  as  the  reciprocal  of  the  3rd  is  to  the  reciprocal  of  the 
4:th.  Thus,  if  a  and  b  are  to  each  other  inversely,  or  recipro- 
cally, as  m  and  n,  a  \b  \:n:m,  or  what  is  the  same  thing, 

«  :  0  : :  —  :-• 
m    n 

04.  A  Continued  Proportion  is  a  succession  of  equal 
ratios,  in  which  each  consequent  is  the  antecedent  of  the 
next  ratio.  Thus,  if  a  :  b  : :  b  :  c  : :  c  :  d  ::  d  :  e.  we  have  a 
continued  proportion. 


OS.  Prop.  1. — In  a  -proportion  the  product  of  the 
extremes  equals  the  product  of  the  means. 

Demonstration.— If  a:h  ::  c:d  then   ad  =  ic.     For  a:h  ::  c:d 

is  the  same  as  £  =  ^ ,  which  cleared  of  fractions  becomes  ad  =  bc. 
0      a 

Q.  E.  D. 

00,  Cor.  1. — The  square  of  a  mean  proportional  equals  the 
product  of  its  extremei^y  and  hence  a  mean  proportional  itself 
equals  the  square  root  of  the  prodtict  of  its  extremes.  For,  if 
a  '.  m  '.:  m  :  d,  by  the  proposition  m^  =  ad.  Whence  ex- 
tracting the  square  root  of  each  member,    m  =  Vad. 

07.  Cor.  2. — Eithi^r  extreme  of  a  proportion  equals  the 
product  if  th'  means  divided  by  the  olh^r  extreme ;  and,  in 
like  manner,  either  mean  equals  the  product  of  the  ewtremes 


288  RATIO,   PROPORTION,   AND   PROGRESSION. 

divided  by  the  other  mean.     For,  if  a  :  b  ::  c  :  d,  ad  =  he. 

^       he  he     ^       ad        ^  ad 

.'.  «  =  — ,  «  =  — ,  0  =z  — ,  and  c  =  ^-  • 
a  d  e  h 


68,  Prop.  2. — //  the  product  of  two  quantities 
equals  the  product  of  two  others,  the  tivo  former  may 
he  made  the  extrem^es,  or  the  means  of  a  proportion, 
and  the  two  latter  the  other  term^s. 

Demonstration. — Suppose  my  =  nx.  Dividing  each  member  by 
ojy,  we  have  —  =  - ;   i.  e.^  m-.x-.'.  n\  y.     In  like  manner  dividing 

by  mn  we  have  -  —  —,  i.  e.,  y:n::x\m. 

Let  the  pupil  determine  how  each  of  the  following  forms 
may  be  deduced  from  the  relation  my  =  nx. 

Given  above. 

By  what  do  you  divide  ? 

Griven  above. 

Dividing  mx  by  each  mem- 

X  TTl 

ber  we  have  -  =  — 
y     n 

By  what  do  you  divide  ? 

How  obtained  ? 


1. 

m  : 

X  : : 

n  :  y. 

2. 

m  : 

n  :: 

X  :  y. 

3. 

y  ' 

n  :: 

X  :  m 

4. 

X  : 

y  :: 

m  :  n. 

5. 

y  ' 

X  :: 

n  :  m. 

6. 

X  : 

m  : : 

y  '  n. 

7. 

71  : 

m  '.: 

y  :  X. 

8. 

n  : 

y  :•• 

m :  X. 

TRANSFORMATIONS  OF  A  PROPORTION. 

6*.9.  Prop.  3. — Proposition  1,  together  with  the  two 
principles  that  such  changes  in  the  terms  of  a  propor- 
tion may  he  made,  as, 

1.  Do  not  change  the  values  of  the  ratios, 

2.  Change  both  ratios  alike, 

are  sufficient  to  determine  in  all  cases  what  transfor- 
mations are  possible  without  destroying  the  proportion. 
That  these  two  principles  are  correct  is  evident  from  the  nature 
of  a  proportion,  as  an  equality  of  ratios. 


PROPOETION.  289 


EXAMPLES. 

MULTIPLES. 

Ex.  1.  li  (I'.h  :\x:y,  prove  that  ma  :mb  I'.xiy. 

Solution. — This  change  does  not  alter  the  value  of  the  first 
ratio,  and  hence  the  equality  of  ratios  remains. 

2.  If  a:b  ::  X  :y  is  7na  :mh  ::  nx :  ny  ?  Why ?  Is  the 
value  of  either  ratio  changed  ?     Why  ? 

3.  If  a\h'.\x\y  is  ma\h  wnix-.yt  Is  the  value  of 
either  ratio  changed  ?     How  ? 

Ans.,  This  change  does  not  destroy  the  proportion, 
because  it  multiplies  both  ratios  by  the  same  quantity. 

CL  X 

4.  It  a  :b  ::  x:y.  18  a: mb  ::  x  : my  ?  or  —  :  b  : :  —  :  v ? 

or  —:b::  x:my? 
m 

6.  If  the  first  term  of  a  proportion  is  multiplied  by  any 
number,  in  what  four  ways  may  it  be  compensated  so  as 
not  to  destroy  the  proportion  ?  Does  multiplying  the  third 
term  by  the  same  number  compensate  ?  Why  ?  Does 
dividing  the  4th  term  ?  Why  ?  Does  dividing  the  second 
term?    Why? 

6.  If  the  third  term  of  a  proportion  is  divided  by  any 
number,  in  what  ways  can  the  change  be  compensated  so  as 
not  to  destroy  the  proportion  ?  Give  the  reason  in  each  case. 


CHANGES  IN  THE  ORDER  OP  TERMS. 

7.  If  a  :b  ::  x  :  y  is  a  i  x  -.:  b  :  y?    How  are  the  ratios 

changed  ? 

Solution.    a:b  ::  x:y  is  the  same  as  r-  =  -  (55).     Now  multi- 

h      y  ^     ' 

plying  each  member  of  this  equality  by  - ,  we  have 
13 


290  RATIO,    PROPORTION,    AlsD  PROGRESSION. 

a  h  X  d 
=-  X  -  =  -  X  - 
0      X      y      X 

«      ^        1-.  1.  • 
or  -  =  - ,  which  18 
X      y' 

a:x  ::i:y. 
Thus  we  see  that  a:i  ::  x:y  is  transformed  into  a:x  ::h:y  by 
multiplying  both  ratios  by  - .     This  does  not  destroy  their  equality 
by  (69,  2). 

00 Cf'.  Hence  we  see  that  we  can  change  the  order  of  the  means  with- 
out destroying  a  j^roportion. 

8.  It  m  :x  ::n:y,  is  n:7n::  y:x?  How  are  the  ratios 
changed  ? 

7Tl  0C71 

Ans.,  Yes.  The  first,  — ,  is  multiplied  bv  -^,  and  the 
second,  -,  is  multiplied  by  — ,  and  — ^  =  ^,  since  nx=my. 

y  Xfh  7fl  71/X 

9.  If  four  quantities  are  in  prcrportion  are  tbey  in  propor- 
tion by  inversion  ?    How  are  the  ratios  changed  ? 

Solution.     a:l) '.'.  c:d  is  the  same  as  t  =  -j,  by  the  definition  of 

o       di 

a  proportion.    Now  dividing  1  by  each  member  of  this  equality  I 

have 

I      d       .... 
-  =  - ,  which  18 
a      c 

h:a::  d:c. 

The  substance  of  this  is  that  if  two  quantities  are  equal,  their 
reciprocals  are  equal. 

00b*  Hence  we  see  that  we  can  take  a  proportion  hy  inversion  with- 
out destroying  it. 

10.  If  3a^  :  W  :  :  6mx  :  lOm^x^,  is  2  :  a^  :  :  6mx  :  &»? 

Solution. — Taking  the  proportion,  and  cancelling  like  factors  from 
both  terms  of  the  same  ratio,  which  does  not  change  its  value,  and 
like  factors  from  both  antecedents,  which  divides  both  ratios  and 
hence  does  not  destroy  the  proportion,  and  like  factors  from  both 
consequents,  which  divides  both  ratios,  and  hence  does  not  destroy 
the  proportion,  we  have 


PROPORTION.  291 

or     fl2 .  ^  .  .  2  :  5mx. 
Whence  changing  the  order  of  the  ratios  2  :  5mx  : :  a' :  &',  and 
finally  changing  the  order  of  the  means  (69,  2),  we  have 
2  :  a'  : :  5mx  :  ¥. 

11.  If  ^03^  :  i%  : :  a^x  :  %,  show  that  J  :  i  : :  a^  :  fa:. 


COMPOSITION    AND    DIVISION. 

12.  m  a  :  b  ::  m  :  n,  show  that  a-\-b  :  a  : :  m  -{- n  :  m. 

Solution.    a:b  ::  m:n  is  the  same  as  ^  =  —  .     Now  add  -  to 

on  0 

the  first  member  and  -  to  the  second,  and  we  have  -  +  -  =  —  +  -, 
n  0      b      n       n 

or  —V-  = ;  that  is  a  +  o  :b  ::  m  +  n  :n, 

h  n 

13.  If  a  :  J  : :  a; :  y,  show  that  a—h  :b  ::  x—y  :  y. 
Suggestion. — Subtract  from  the  first  ratio  -=^,  and  from  the  sec- 

ond^. 

y 

14.  If    m  \n  :\x  \  y,    show    that    m-\-n  \m—n  :'.x-\-y 
IX— y. 

Suggestion. — By  the  method  of  Ex.  12,  I  have  =  — -^ 

and  by  that  of  Ex.  13,  ^11^  =  ^=^. 

n  y 

Dividing  the  former  by  the  latter,  I  have  =  — -\ 

m-n      x—y^ 
that  is,  m-^n'.m—n  ::  x-\-y:x—y. 

SOc,   Hence  we  see  that  a  proportion  may  le  taken  by  composition^ 
or  by  division^  or  by  both  at  once,  without  destroying  it. 

16.  If    ia—x  :  Ja  +  ic  ::  b—y  :  b  +  y,    show   that    2x  :  y 
::  a  :b. 

16.  li  a  :b  ::  X  :  y,  does  it  follow  that  a—y  :  b—x  ::a:x? 

Ans.,  No. 


RATIO,   PROPORTION^,   AN^D   PROGRESSIOl?". 

17.  From  a\l\\x\y,  prove  that  ma+nb  :  ma—nh 
::mx-\-ny  :  mx—ny.    Also  that  if  a\l  \\c\d,  and  m  :  n 

y  J         T  a     b      c    d 

\ix  :  y,  am-.bnw  ex  \  ay,  and  —:-::-:-• 

m    n      X    y 

IS.  li  a  :  b  ::  X  :  y  is  a^ :  b^  ::  x^ :  y^?  Is a^  :  bn  ::  x^  :  yn? 
[s  Va  :  Vb  : :  V^^  :  Vy  ?  Is  a^  :b^  ::  x^  :  y^?  Is  a^  :  b"^ 
::  x»^  :  y^  whether  m  is  integral  or  fractional,  positive  or 
negative  ?  Why  is  it  that  the  ratios  remain  equal  in  each 
case  ?    How  are  they  changed  ? 


MISCELLANEOUS. 


19.  lia  \b  ::  c  :  d,  show  that  ?na  +  tz^  '.pa-\-qb  ::  mc-\-nd 
'.pc-\-qd, 

'    Suggestion. — The  ratios  to  be  compared  when  reduced  to  a  C.  D. 
acmjj  +  bcnp  +  admq  +  bdnq  acmp  +  bcmqi-adnp  +  Mnq 

'  {ap-\-dq){cp  +  dq)  {ap-^l)q){cp  +  dq) 

Now  from  the  given  proportion  we  learn  that  ad  =  he.  Therefore, 
exchanging  them  in  the  two  middle  terms  of  the  first  ratio,  the 
ratios  become  identical. 

This  may  also  be  shown  as  follows :  Multiplying  antecedents  by 
m  and  consequents  by  n,  ma  :7il)  ::  mc:  nd.  By  composition 
ma-{-nb:ma  ::  mc  +  nd:mc,  or  multiplying  both  ratios  by  wi, 
Tna  +  nb-.a-.imc  +  ndic.  By  changing  the  places  of  the  means 
ma-\-7il):mc-\-nd  ::  a:c.  In  like  manner  it  may  be  shown  that 
pa  +  qh:pc  +  qd  ::  a:c.  .'.  ma  +  nhimc  +  nd  ::  pa  +  qb  :pc  +  qd,  or 
ma-{-ni:2>a-\-qb  ::  n),c  +  nd:  pc-i-qd.  The  student  should  give  the 
reason  why  each  step  does  not  vitiate  the  proportion,  according 
to  (69). 

20.  lt{a  +  b-}-c-i-d){a  —  b  —  c-j-d)  =  (a—b  +  c—d) 
{a-\-b—c—d)  prove  that  a  :  b  \:  c  \  d. 

Suggestion. — Performing  the  operations  and  reducing,  2ad—%bc 

==  —%ad-\-%l)c,  or  ad  =  5c.    Whence  ^  =  -^ ,  or  a:b::  c:d, 
'  0      a 


PROPORTION.  293 

This  may  also  be  proved  by  writing  according  to  Prop.  2,  a+J 
+  c-i-d :  a—b-\-c—d  : :  a-{-b—e — d  :  a—h-c  +  d.  Comparing  the  sum 
of  each  antecedent  and  its  consequent  with  their  difference,  2a 
-\-2c:2b  +  '-Zd::2a  —  2c:2b—2d.  Whence  a +  c:  a-c  ::  b-\-d:h—d. 
Repeating  the  same  processes  we  have  a:b::  c:d. 

21.  If  — - —  =  h,  show  that  a—x  :2a  ::  2b  :  a  +  x.   Pro- 

duce  other  forms  of  proportion  from  the  given  relation. 
How  many  can  be  produced  ? 

22.  If  r  =  s\/|,  show  that  r  :  s  : :  1  :  ^2. 

23.  If  (a-\-xY  :  {a—xY  \'.x-\-y  :x—y,  show  that  a:x 
::  \/2a—y  :  \/y. 

Solution.  a'  +  2aa;4-aj'  \a^—2m-^x^  ::  x+y:x—y^ 

2a'  +  2x'»  :^ax.:2x:  2y,        a'-k-x^ :  ckV  :  2«  :  y, 
a?  \^  \ :  2a— y  :  y,   .'.  a\x\:  ^2a—y :  ^y. 
Let  the  student  give  the  reasons. 

24.  li  a  :  h  :'.  c  :  d  ::  e  :  f '.:  g  '.  h  w  i  '.  k,  etc.,  show  that  - 
{a-\'C-\-e-\-g-\-i+,etc.)  :{h-\-d-^f-\-h-\-k-\-,Qic.)  ::a:b,or 
c  :  d,  or  e  :f,  etc.    That  is,  in  a  series  of  equal  ratios,  the 
sum  of  all  the  antecedents  is  to  the  sum  of  all  the  conse- 
quents, as  any  antecedent  is  to  its  consequent. 

Solution.    zr  =  rorab  =  ba,        -  =  -^  or  ad  =  be. 
b      b  b      d 

a       e         ,,  a      g         .,  a       i         i       i-     . 

,-  =  -  or  «/  =  &e,         =^  =  Y  or  ah  =  bg,        -  =  -  or  ak  =  bi,  etc. 

b         J  b         h  OK 

Adding,  a(&4-<^+/+^  +  *  +  ,  etc.)  =  6(a  +  c  +  e  +  ^  +  i  +  ,  etc.); 
whence  {a-\-c-\-e  +  g-\-i  +  ^  etc.)  :  (&  +  (i+/+A  +  ^  +  ,  etc.)  : :  a  :  &  or 
(since  a:b  =  c:d^  etc.),  as  c:d::  e:f,  etc. 

25.  Four  given  numbers  are  represented  by  a,  h,  c,  dy 
what  quantity  added  to  each  will  give  sums  which  are  pro- 
portional. .  he— ad 

a—o—c+d 

26.  If  four  numbers  are  proportionals,  show  that  there  is 
no  number  which,  being  added  to  each,  will  leave  the 
resulting  four  numbers  proportionals. 


Pf\OGf\ESSiON 


70,  A  Progression  is  a  series  of  terms  which  increase 
or  decrease  by  a  common  difference,  or  by  a  common  mul- 
tiplier.* The  former  is  called  an  Arithmetical,  and  the 
latter  a  Geometrical  Progression. 

A  Progression  is  Increasing  or  Decreasing  according  as 
the  terms  increase  or  decrease  in  passing  to  the  right.  The 
terms  Ascending  and  Descending  are  used  in  the  same  sense 
as  increasing  and  decreasing,  respectively. 

In  an  increasing  Arithmetical  Progression  the  common 
difference  is  added  to  any  one  term  to  produce  the  next 
term  to  the  right ;  and  in  i  decreasing  progression  it  is 
subtracted. 

In  an  increasing  Geometrical  Progression  the  constant 
multiplier  by  which  each  succeeding  term  to  the  right  is 
produced  from  the  preceding  is  more  than  unity  ;  and  in  a 
decreasing  progression  it  is  less  than  unity.  This  constant 
multiplier  in  a  Geometrical  Progression  is  called  the  Batio 
of  the  series. f 

71.  The  character,  . .,  is  used  to  separate  the  terms  of 
an  Arithmetical  Progression,  and  the  colon,  :,  for  a  like 
purpose  in  a  Geometrical  Progression. 

*  This  is  the  common  use  of  the  term.  It  Is  aleio  used  to  include  what  is  called 
a  Harmonical  Progression. 

t  This  is  an  unfortunate  use  of  the  term  Eatio,  inasmuch  as  it  is  at  variance  with 
its  use  in  proportion.  To  harmonize  the  uee  of  the  term  in  proportion,  with  this 
use,  may  have  led  some  writers  to  define  ratio,  as  used  in  proportion,  as  the  quo- 
tient of  the  consequent  divided  by  the  antecedent.  But  the  definition  has  neither 
logic  nor  the  common  usage  of  authors,  English  or  Continental,  to  support  it.  The 
French  use  rapport  in  proportion,  and  raisan  in  progression. 


ARITHMETICAL    PROGRESSION.  296 

Illustrations. 

1 .  .3.  .5.  .7,  etc.,  is  an  Increasing  Arithmetical   Progression  with  a 

common  difference  2.  or  +2. 

15.  .10.  .5.  .0.  .—5,  etc.,  is  a  Decreasing  Arithmetical  Progression 

with  a  common  difference— 5. 

a.  .a  ±d.  .a±2d.  .a±Sd,  etc.,  is  the  general  form  of  an  Arith- 
metical Progression,  d  being  the  com- 
mon difference. 

2  :  4  :  8  :  16,  etc.,  is  an  Increasing   Geometrical  Progression  with 

ratio  2. 

12  :  4  : 1 : 1 :  TjSiy,  etc.,  is  a  Decreasing  Geometrical  Progression  with 

ratio  J. 

a:ar:ar^ :  ar* :  ar*,  etc.,  is  the  general  fonn  of  a  Geometrical  Pro- 
gression, r  being  the  ratio,  and  greater 
or  less  than  unity,  according  as  the 
series  is  increasing  or  decreasing. 


72.  There  are  Five  Tilings  to  be  considered  in  any  pro- 
gression; viz.,  the  tirst  term,  the  last  term,  the  common 
difference  or  the  ratio,  the  number  of  terms,  and  the  sum  of 
the  series,  either  three  of  which  being  given  the  other  two 
can  be  found,  as  will  appear  from  the  subsequent  discussion. 


7 


75.  Prop.  1. — The  formula  for  finding  the  nth,  or 
last  term  of  an  Arithm,etlcal  Progression;  or,  more 
properly,  the  formula  expressing  the  relation  between 
the  first  term,  the  nth  term,  the  com^mon  difference, 
and  the  nurriber  of  terms  of  such  a  series  is 

I  =^  a  -\-  (n  —  1)  d, 
in  which  a  is  the  first  term,  f?  the  common  difference,  n  the 
number  of  terms,  and  li\\Qni\\  orla.stterm,  d  being  positive 
or  negative  according  as  the  series  is  increasing  or  decreasing. 


296  RATIO,   PROPORTION,   AND   PROGRESSION. 

Demonstration. — According  to  the  notation,  the  series  is 
a.  .a  +  d.  .a-\-2d.  .a  +  dd.  .a  +  4:d.  .a  +  5d^  etc. 
Hence  we  observe  that  as  each  succeeding  term  is  produced  by  add- 
ing the  common  diflerence  to  the  preceding,  when  we  have  reached 
the  7it\i  term,  we  shall  have  added  the  common  difference  to  the  first 
term  n—1  times ;  that  is,  the  nth  term,  or  1=  a  +  {n—l)d.    q.  e.  d. 

Scholium. — As  this  formula  is  a  simple  equation  in  terms  of  a,  l, 
n,  and  d,  any  one  of  them  may  be  found  in  terms  of  the  other  three. 


74.  Prop.  2. — The  formula  for  the  sum  of  an 
Arithmetical  Progression,  i.  e.,  expressing  the  rela- 
tion between  the  sum  of  the  series,  the  first  term,  last 
term,  and  nurrhber  of  terms  is 

ra  +  li 

'  =  b^h 

s  representing  the  sum  of  the  series,  a  the  first  term,  I  the 
last  term,  and  n  the  number  of  terms. 

Demonstration. — If  I  is  the  last  term  of  the  progression,  the  term 
before  it  is  l—d,  and  the  one  before  that  I— 2d,  etc.    Hence,  as  a.  .a 

+  d.  .a  +  2d.  M  +  Sd ^,  represents  the  series,  I.  .l—d.  .l—2d.  .1 

— 3<Z a,  represents  the  same  series  reversed.     Now,  the  sum 

of  the  first  series  is 

s  =  a+  {a-\-  d)  +  {a  +  2'J)  -\- {I— 2d)  +  (l—d)  + 1; 

nnd  reversed,   s  =  1+  (l—d)  +  Q—2d)  +     {a-ir2d)-\-{a  +  d)  +  a. 

Adding,  2«  =  {a  +  l)  +  {a-^l)  +  {a  +  l)+ {a  +  l)  +  {a  +  T)  +  {a  +  l). 

If  the  number  of  terms  in  the  series  is  n^  there  will  be  n  terms  in 
this  sum,   each   of  which   is   (a  +  Z) ;    hence    28  =  (a  +  l)n,   or 

a  +  r 


ra  +  n 


Q.  E.  D. 


Scholium. — This  formula  being  a  simple  equation  in  terms  of  «, 
a,  Z,  and  n,  any  one  of  the  four  can  be  found  in  terms  of  the  other 
three. 

7S.  Cor.  1. — Formulas 

(1)  1  r=  a  +  (n  —  1)  d,        and 

(2)  ^  —  \  ~~J~  P^  being  two  equations 

hetioeen  the  five  quantities,  a,  1,  n,  d,  aiid  s,  any  ttvo  of  these 
five  can  befotmd  in  terms  of  the  other  three. 


ARITHMETICAL    PROGRESSION.  297 

[Note. — It  is  not  considered  worth  while  to  make  separate  cases 
out  of  the  diiferent  problems  which  arise  in  the  progressions,  or  to 
cumber  the  memory  with  the  multiplicity  oi' /(/rmulaa  which  can  be 
deduced  from  the  two  fundamental  ones,  but  rather  that  these  should 
be  fixed  in  memory,  and  their  use  clearly  understood.] 

EXAMPLES. 

Ex.  1.  The  first  term  of  an  A.  P.  is  2,  and  the  common 
difference  3,  what  is  the  11th  term  ?  What  the  sum  of  the 
series  ? 

Solution. — In  the  first  case  there  are  under  consideration  the  first 
term,  a  —  2,  the  common  difference,  d—3,  the  number  of  terms, 
n  =  11,  and  the  last  term,  which  is  the  thing  required.  The  relation 
between  these  is  given  in  I  =  a-\-{n—l)d;  in  which  by  substituting 
the  given  values  there  results  I  =  3  +  (11  — 1)3  =  32.     In  the  second 

case  the  formula  «  =    — —  \n  gives  the  relation,  in  which  by  sub- 

— - —    11  =  187. 

2.  The  first  term  of  an  A.  P.  is  8,  the  last  term  203,  and 
the  common  difference  5,  what  is  the  number  of  terms  ? 
What  the  sum  of  the  series  ?        Ans.,  n  =  40,  s  =  4220. 

3.  The  first  term  of  an  A.  P.  is  8,  the  last  term  203,  and 
the  number  of  terms  40,  what  are  the  common  difference 
and  the  sum  ? 

4.  The  last  term  is  1,  the  sum  1717,  and  the  number  of 
terms  34,  what  are  the  first  term  and  the  common  differ- 
ence? 

Suggestion. — The  equations  are  1  =  a  +  (34— !)(?,  and  1717 
=  (-^ )  34)  fr""i  which  to  find  a  and  d.    a  =  100,  and  d=  -^S. 

5.  What  is  the  sum  of  the  numbers  1, 2, 3, 4,  etc.,  to  1000  ? 

6.  The  first  term  of  an  arithmetical  progression  is  1,  and 
the  number  of  terms  23,  what  must  be  the  common  differ- 
ence  that  the  sum  of  all  the  terms  may  be  100  ?  What  the 
last  term  ? 


298  RATIO,   PltOPvJllTlON,    AXD   PKOGRESSIOK. 

7.  If  tlie  lirst  term  of  an  arithmetical  progression  is  100, 
and  the  number  of  terms  21,  what  must  the  common  differ- 
ence be  that  the  sum  of  the  series  may  be  1260  ?  What  the 
last  term?  A^is.,   —4. 

8.  Two  persons,  A  and  B,  start  from  the  same  place 
together,  and  travel  in  the  same  direction.  A  goes  40  miles 
per  day ;  B  goes  20  miles  the  first  day,  and  increases  his 
rate  of  travel  f  of  a  mile  per  day.  How  far  will  they  be 
apart  at  the  end  of  40  days,  and  which  will  be  in  advance  ? 

Ans.,  A  will  be  215  miles  in  advance  of  B. 

9.  The  first  term  of  an  arithmetical  progression  is  —7, 
the  common  difference  —7,  and  the  number  of  terms  101, 
what  is  the  sum  of  the  series  ?  Ans.,  —36057. 

76,  Cor.  2. — The  formula  for  inserting  a  given  number  of 

1— a 

arithmetical  means  between  two  given  extremes  is  d  := r  , 

^  m  +  1 

in  which  m  represents  the  number  of  means.     From  this  d, 

the  common  difference,  being  found,  the  terms  can  readily  be 

written. 

Demonstration. — If  a  is  the  first  term  and  I  the  last,  and  there 
are  m  terms  between,  or  m  means,  there  are  in  all  w  +  2  terms. 
Hence  substituting  in  the  formula  I  =  a  +  {n—l)d  for  n,  m  +  2,  we 

have  l  =  a+  (m  +  l)d.    From  this  d  = -.     Q.  e.  d. 

'  m  +  1 

10.  Insert  8  arithmetical  means  between  3  and  21. 

Series,  5.  .7.  .9.  .11.  .13.  .15.  .17.  .19. 

11.  Insert  3  arithmetical  means  between  -J  and  J. 

Series,  i-.^-.U- 

12.  What  is  the  nth  term  of  the  series  1..3..5..7.., 
etc.?  Ans.,  2n— 1. 

13.  What  is  the  sum  of  n  terms  of  the  series  1 . .  3 . .  5 . .  7 
..,  etc.?  Ans,,  n\ 


GEOMETRICAL    PKOGRESSIOK.  290 

14.  If  a  body  falling  to  the  eai'th  descends  a  feet  the 
first  second,  3a  the  second,  ba  the  third,  and  so  on,  how 
far  will  it  fall  during  the  ^th  second  ?       Ans.,  (2i  —  l)a. 

15.  If  a  body  falling  to  the  earth  descends  a  feet  the  first 
second,  3a  the  second,  5a  the  third,  and  so  on,  how  far  will 
it  fall  in  t  seconds  ?  Ans.,  aP. 

16.  A  debt  can  be  discharged  in  a  year  by  paying  $1  the 
first  week,  $3  the  second,  $5  the  third,  and  so  on  ;  required 
the  last  payment  and  the  amount  of  the  debt. 

Ans.,  Last  payment,  $103 ;  amount,  $2704. 

17.  A  person  saves  1270  the  first  year,  $210  the  second, 
and  so  on.  In  how  many  years  will  a  person  who  saves 
every  year  $180  have  saved  as  much  as  he?  A7is.,  4. 

18.  A  board,  2^  inches  wide  at  the  narrow  end,  and  10 
feet  long,  increases  in  width  Ij^  inches  for  every  foot  in 
length  ;  what  is  the  width  of  the  wide  end  ?     Ans.,  17^  in. 

19.  If  100  oranges  are  placed  in  a  line,  exactly  2  yards 
from  each  other,  and  the  first  2  yards  from  a  basket ;  what 
distance  must  a  boy  travel,  starting  from  the  basket,  to 
gather  them  up  singly,  and  return  with  each  to  the  basket  ? 

Ans.y  11  mi.  3  fur.  32  rd.  4  yd. 

[Note.— For  other  examples  involving  the  principles  of  Arith- 
metical Progression,  see  Problems  after  Quadratics,  and  also  the 
subject  of  Interest.] 


II  II    II    II   II    II  ^ 


77.  Prop.  1. — The  formula  for  finding  the  nth,  or 
last  teimv  of  a  Geometrical  Progression;  or,  more 
properly,  the  formula  expressing  the  relation  between 
the  first  term,  the  nth  term,  the  ratio,  and  the  number 


300  KATIO,  PROPORTION,  AND  PROaRESSlON. 

of  terms  of  such  a  series  is  1  =  ar^-^,  in  which  1  7.5  the 
last,  or  wth  term,  a  the  first  term,,  r  the  ratio,  and  n  the 
number  of  term^s. 

Demonstration. — Letting  a  rei^resent  the  first  term  and  r  the 
ratio,  the  series  is  a  :  ar :  ar^ :  ar^ :  ar* :  etc.  Whence  it  appears  that 
any  term  consists  of  the  first  term  multiplied  into  the  ratio  raised  to 
a  power  whose  exponent  is  one  less  than  the  number  of  the  term. 
Therefore  the  n\\\  term,  or  Z  =  ar''-'^.     q.  e.  d. 

78.  Prop.   2. — The    formula    for    the  sum  of   a 

Geometrical  Progression,  or  expressing  the  relation 
between  the  sum  of  the  series,  the  first  term,  the  ratio, 
and  the  nuinher  of  terms  is 

ar^  —  a 

in  which  s  represents  the  sum,  a  the  first  term,  r  the  ratio, 
and  n  the  number  of  terms. 

Demonstration. — The  sum  of  the  series  being  found  by  adding 

all  its  terms,  we  have, 

8=a-\-ar-\-ar^+ar^-\ ar'^~^  +  ar""-^  ■\- ar'^-\  and  multiplying 

by  r,  rs  =       ar  -f  ar^  +  ar^  -\ a?*"-^  +  ar"~^  +  ar"~'  +  ar". 

Subtracting  rs—  8=^  a/r'^—a^  or 

...  ,  ar^—a 

(/•— l)s  =  arn—a^  and  s  =  — - .     q.  e.  d. 

r — 1 

Suggestion. — The  student  will  notice  that  multiplying  the  first 
series  by  ?•,  and  placing  the  terms  of  the  product  under  the  like 
terms  of  the  series,  simply  moves  each  term,  when  multiplied,  one 
place  to  the  right,  so  that  however  many  terms  there  may  be  in  the 
series,  each  will  have  a  similar  one  in  the  product  except  the  first 
term,  a ;  and  each  term  in  the  product  will  have  a  similar  one  in 
the  series,  except  the  last  one,  ar". 

79.  Cor.  1. — Formulas 

(1)  1  =  ar'^-i,     and 

(2)  s  = —   hei7ig  two  equations 

between  the  five  quantities,  a,  1,  r,  n,  and  s,  are  sufficient  to  de- 
termine any  two  of  them  when  the  others  are  given. 


GEOMETRICAL    PROGRESSION.  301 

80.  Cor.  2.— Since  1  =  ar»-i,  Ir  =  ai*'*,  which  svbstUuted 
in  (2)  gives  s  = r;  which  formula  is  often  convenient, 

EXAMPLES. 
Ex.  1.  The  first  term  of  a  geometrical  progression  is  2, 
the  ratio  3,  and  the  number  of  terms  6.     What  are  the  last 
term  and  the  sum  of  the  series  ? 

nrn d 

Ans.,  l  =  2'd^  =  486,  s  =  ^:— ^  =  728. 

2.  The  last  term  of  a  geometrical  progression  is  62500, 
the  ratio  5,  and  the  number  of  terms  7.  What  are  the  first 
term  and  the  sum  of  the  series  ? 

Ans.,  a  =  4,  and  s  =  78124. 

3.  By  saving  1  cent  the  first  week,  2  cents  the  second 
week,  4  cents  the  third  week,  and  so  on,  doubling  the 
amount  every  week,  how  much  is  saved  the  last  week  of  the 
year  ?  Ans.,  $22,517,998,136,852.48. 

Suggestion. — This  problem  requires  us  to  raise  2  to  the  5l8t 
power.  This  is  readily  effected  thus :  the  3rd  power  of  2  is  8. 
The  3rd  power  multiplied  by  the  3rd  power  gives  the  6th  power ; 
hence  8  x  8  =  64  is  the  6th  power.  In  like  manner  64  x  64  =  4096 
is  the  12th  power,  and  4096  x  4096  =  16777216  is  the  24th  power; 
and,  finally,  16777316  x  16777216  x  8  is  the  5l8t  power. 

4.  What  is  the  sum  of  10  terms  of  the  series  8:4:2:1 
:  J,  etc.  ? 

•ft.-) 


a(r*-l)  ^    V2"  /  _  ^,  ^  2^°-l  _  1023 

r-1     ~  1 


Suggestion.    «  =     ^    .,     = ^ =  2*  x 


2 

=  — — -  =  1554.     Also,  I  =  ar*-'  =  —  =  —  =  A-.  In  such  cases  the 
64  ^*  '  2"       2"      " 

ingenious  student  will  avoid  unnecessary  multiplications  by  sup- 
pressing factors. 

5.  If  4  is  the  first  term,  324  the  last,  and  5  the  number 

of  terms,  what  is  the  ratio  ? 


302  EATIO,   PROPORTION",   ANB   PROGRESSION. 

81,   Cor.  3. — The  formula  for  inserting  m  geometrical 

m+l/i 

means  between  a  and  lis  r  ^  \/  ~' 

6.  Insert  5  geometrical  means  between  3  and  192. 

The  ratio  is  2. 

7.  Find  4  geometrical  means  between  ^  and  ^. 

Result,  The  ratio  is  l-j,  whence  the  series  becomes 

l._l_._l_     _J_         1        1 

^'2^3^21.3^2^.3^' 2*.  3^*3* 

Scholium. — This  and  many  other  problems  in  Geometrical  Pro- 
gression, are  more  readily  solved  by  means  of  logarithms.  Many 
also  require  a  knowledge  of  quadratic  equations,  and  even  of  the 
higher  equations.  Some  farther  illustrations  will  be  given  in  their 
proper  place,  especially  in  treating  the  subject  of  Interest. 

S2,  Cor.  4. — The  formula  for  the  sum  of  an  Infinite  De- 
creasing Geometrical  Progression  is  s 


1— r 


Demonstration. — Since  in  a  decreasing  progression  the  ratio  is 
less  than  unity,  the  last  term,  ar''-^  is  also  less  than  the  first  term, 

and  numerator  and  denominator  of  the  value  of r,  both  become 

r — 1 

negative.   Hence  it  is  well  enough  to  write  the  formula  for  the  sum  of 

such  a  series  s  =  :j — -  ,  that  is,  change  the  signs  of  both  terms  of 

the  fraction.  Now,  if  the  terms  of  a  series  are  constantly  decreas- 
ing, and  the  number  of  terms  is  infinite,  we  can  fix  no  value,  how- 
ever small,  which  will  not  be  greater  than  the  last,  or  than  some 
term  which  may  be  reached  and  passed.  Hence  we  are  compelled 
to  call  the  last  term  of  such  a  series  0,  which  makes  the  formula 
a 

S  =    . .      Q.  E.  D. 

1—r 

Schoh'um.—  Decimal  Repetends  afford  illustrations  of  such  series. 
Thus  .833  +  ,  is  A  +  Tiir+T^(nr  +  ?  etc.,  to  infinity.  Again,  .5434343  + 
is,  -^^^-fthe  series  ,^3^+^^4|^  +  ^^^^^4|^^^  +  ,  etc.,  to  infinity. 


GEOMETRICAL    PROGRESSION.  303 

8.  Find  the  sum  of  the  series  1  :  J  :  J  :  etc.,  to  infinity. 

Suniy  2. 

9.  Required  the  sum  of  the  series  1,  i,  i,  —  to  infinity. 

Sum,  1\. 

10.  Find  the  value  of  .1212  -  -  -  to  infinity.    Sum,  -^, 

11.  Find  the  value  of  .2333,  etc.,  to  infinity.  Su7)iy  ^. 

12.  Find  the  value  of  .3411111,  etc.,  to  infinity. 

Sum,  m- 

13.  Find  the  value  of  .323232,  etc.,  to  infinity.  Sum,  ||. 

14.  Find  the  value  of  .20414141,  etc.,  to  infinity. 

Sum,  |«i. 

15.  Suppose  a  body  to  move  eternally  in  this  manner ; 
viz.,  20  miles  the  first  minute,  19  miles  the  second  minute, 
IS^ij^  the  third,  and  so  on  in  geometrical  progression.  What 
is  the  utmost  distance  it  can  reach?         Ans.,  400  miles. 

16.  What  is  the  distance  passed  through  by  a  ball,  before 
it  comes  to  rest,  which  falls  from  the  height  of  50  feet,  and 
at  every  fall  rebounds  half  the  distance  ?  Ans.,  150. 

17.  In  the  preceding  problem,  suppose  the  body  falls 
16^^  feet  the  first  second,  3  times  as  far  the  next  second, 
and  5  times  as  far  the  third  second,  and  so  on,  how  long 
will  it  be  before  it  comes  to  rest  ? 

Ans.,  ^  V579  (4  +  3  a/2)  =  10.27657+  seconds. 

Suggestion.— To  fall  50  ft.  takes  4/  -— —  sees.     To  rebound  25 

'     1^ 

/  25 
takes  Y  Ygx  ««».  and  to  fall  25  the  same  time.   To  rebound  12^  ft. 

takes  y  j^  and  the  same  to  fall.      Hence   the  series  giving 


the  time  is  T  =  a/ 


50 


^  j     /  25  1/25        1/25  ,  .  ^  .       ) 


Va 


304 


RATIO,   PEOPORTION,   AKD   PKOGRESSIOK. 


SYNOPSIS. 


CO 

(f> 

LU 
CC 
O 
O 
CC 
Q. 


^^ 


a. 
o 

QC 

a. 


O 

h- 

o 
o 

fiC 

CL 

< 

CC 


Definitions. 


Definitions. 


Fundamen- 
tal Propo- 
sitions. 


Transfor- 
mations 


^  Definitions. 


A.  P. 


G.  P. 


f  Ratio.— Terms  of.— Antecedent.— Consequent.— 
t  Conplet. 

Direct.— Inverse.— Greater  Inequality,  less. 
L  Compound  ratio.— Duplicate,  sub-duplicate,  etc. 

Sign  of. 

Cor.— Changes  in  terms  of. 

'  Proportion.— Extremes.— Means. 

Mean  proportional.— Third  proportional. 

Inversion. — Alternation. — Composition.— Division. 
.  Inverse  or  reciprocal  proportion.— Continued. 

■  Prop.  1.—  Cor.  l.—Cor.  2. 
Prop.  2. 

Prop.  3.— Principles  on  which  transformations  are 
made. 

Equi-mnltiples.— Why  proportion  not  destroyed. 
Chg.  in  order  of  terms.—      "  "  " 

Composition,  or  division.—  "  "  " 

Involution,  or  evolution.—  "  "  *' 

Progression.—  Arithmetical.— Geometrical. 
Increasing,  or  ascending.— Decreasing,  or  descend- 
ing. 
Common  difference,  positive,  negative. 
Ratio,  greater  than  1,  less  than  1. 

Sign  of  Arithmetical  Progression. — Of  Geometri- 
cal. 

Five  things.— Given,  required. 

Two  ftindamental  formulas.— Produce  them.— To 
insert  means. 

Two   fundamental  formu-  j  To  insert  means. 

las.— Produce  them.    I  Sum  of  infinite  series. 


Test  Questions.- -Give  the  various  changes  which  can  be  made 
upon  the  terms  of  a  ratio  and  tell  how  the  ratio  is  affected,  and 
Why  f  State  the  various  transformations  which  can  be  made  upon 
a  proportion  without  destroying  it,  and  Gim  the  reason  in  each  case. 
Produce  the  two  fundamental  formulas  of  Arithmetical  Progression. 
Also  of  Greometrical. 


APPLICATIONS.  305 

APPLICATIONS. 

[Note.— Teacher  and  pupil  should  bear  in  mind  that  the  object  of 
this  section  is  to  teach  the  properties  of  Ratio  and  Proportion ; 
hence  all  the  operations  should  be  performed  upon  the  proportion. 
The  proportion  should  be  kept  in  the  form  of  a  proportion,  and  not 
reduced  to  an  equation.  ] 

Ex.  1.  Divide  60  into  two  parts  which  are  to  each  other 
as  2  :  3. 

Suggestion. — Letting  x  and  60— x-  be  the  parts,  x  :  60— a; : :  2  :  3. 
Hence  x  :  60  : :  2  :  5,  or  aj :  24  : :  2  :  2 ;  and  x=24.  The  pupil  should 
give  the  reason  for  each  transformation.  What  is  the  first  trans- 
formation ?  Composition.  Why  does  it  not  destroy  the  propor- 
tion ?  What  the  second  transformation  ?  Why  does  it  not  destroy 
the  proportion  ? 

2.  A  boy  being  asked  his  age  said :  John,  who  is  18,  is 
older  than  I;  but,  if  you  add  to  my  age  \  of  it,  and  from 
this  sum  subtract  \  of  my  age,  the  result  will  be  to  John's 
a^e  as  10  :  9.    How  old  was  the  boy  ?    Verify. 

Operation.  a;+^a;— i«:  18  ::  10  :  9, 

fx  :  18  : :  10  :  9, 
a;:18::    8:9, 
a? :  18  : :  16  :  18, 
.-.  x  =  16. 
Let  the  pupil  tell,  in  each  instance,  just  what  the  transformation 
is,  and  why,  according  to  (69),  the  proportion  is  not  destroyed. 

Verifloatlon.  16  + J/— V  =  ^^^  which  is  to  18  as  10  is  to  9,  the 
ratio  being  J^. 

3.  Two  brothers  being  asked  their  ages,  the  younger  re- 
plied, my  age  is  to  my  brother's  as  2  to  3 ;  and  if  you  add  18 
to  mine  and  2  to  his,  the  sums  will  be  as  3  to  2.  What 
were  their  ages  ? 

Suggestion. — To  solve  with  om  unknown  quantity  we  may  repre- 
sent the  younger  brother's  age  by  2a;  and  the  older's  by  3a;. 
Then  2j;  +  18  :  3x  +  2  ::  3  :  2; 

Whence      2a;+18  :  27a;+18  ::  3: 18, 

2x  +  18:25a;::  3:15::  1:5. 
2ic+18:«::  5:1, 


306  PROPORTION. 

2a5  +  18:2iK::  5:3, 
18 :  2a; : :  3  :  3, 
9  :  X  : :  9  :  6. 
.*.  X  =  Q,  2x  =  12  and  3a;  =  18. 
[^o\e. — True,  it  gives  a  somewhat  shorter  solution  of  this  example 
to  put  the  first  proportion  immediately  into  the  equation  4a; +  36 
=  9a;  +  6,  whence  ox  =  30  and  x  =  Q.     But  the  object  is  to  become 
familiar  with  the  properties  of  a  proportion.] 

4.  A  man's  ao:e  when  he  was  married  Avas  to  his  wife's  as 
3  to  2 ;  but  after  4  years,  his  age  was  to  hers  as  7  to  5. 
What  were  their  ages  when  they  were  married  ? 

Suggestion. — This  may  be  solved  with  one  unknown  quantity, 
like  the  preceding,  and  that  is  the  more  elegant  way.  We  may  also 
use  two.  Thus,  a;:  y  : :  3  :  3,  and  a;+4  :  y  +  4  : :  7  :  5.  From  the 
former  a; :  fy  : :  3  :  3  .  •.  x  =  §y.  Substituting  in  the  latter  |y 
+  4:y  +  4  ::  7:5.  Whence  Sy  +  S  :2y  +  S  ::  7:5,  y:2y  +  8::  3:5, 
2y:2y  +  S  ::4:5,  3?/ :  8  : :  4  : 1,  and  i/ :  1  : :  16  :  1.  .'.  y  =  16,  and 
a;  =  |y  =  34. 

5.  A  man  is  now  25  years  old  and  his  brother  is  15.  How 
many  years  before  their  ages  will  be  as  5  to  4  ?     Verify. 

6.  A  man  has  two  flocks  of  sheep,  each  containing  the 
same  number ;  from  one  he  sold  80  and  from  the  other  20, 
when  the  numbers  in  the  flocks  were  as  2  to  3.  -  How  many 
were  there  in  each  flock  in  the  first  place  ?     Verify. 

7.  It  is  known  to  every  one  that  a  small  body  near  the  eye 
hides  a  large  one  farther  off ;  and  it  is  a  principle  in  optics 
so  nearly  axiomatic  that  we  will  take  it  for  granted,  that,  in 
order  to  have  the  smaller  body  just  cover  the  larger  their 
distances  from  the  eye  must  be  in  proportion  to  their  breadths, 
or  lengths.  From  this  some  very  pleasing  calculations  can  be 
made.     The  pupil  may  make  the  following : 

1st.  The  breadth  of  a  man's  thumb  is  about  1  inch,  and 
he  can  readily  hold  it  at  2  feet  from  his  eye ;  how  far  off  is 
the  man  who  is  5  feet  8  inches  high,  when  the  breadth  of 
the  man's  thumb  at  2  feet  from  his  eye  just  covers  the  man  ? 

Ans.,  136  feet. 


APPLICATIONS.  307 

2nd.  Wishing  to  know  approximately  the  height  of  the  top 
of  a  steeple  from  the  ground,  I  found  that  my  hand,  which 
is  4  inches  wide  near  the  thumb,  when  held  2  feet  from  my 
eye,  just  covered  the  height  of  the  steeple  at  a  distance  of 
240  paces  of  3  feet  each.  What  was  the  height  of  the  top  of 
the  steeple  from  the  ground  ?  Ans.y  120  feet. 

8.  What  number  is  that  to  which  if  1,  5,  and  13  be  sev- 
erally added,  the  second  sum  will  be  a  mean  proportional 
between  the  other  two  ?     Verify. 

9.  What  ii umber  is  that  whose  |  increased  by  2  is  to  its  | 
diminished  by  1,  as  6  is  to  2J?  A7is.,  30. 

10.  The  number  of  acres  a  farmer  planted  with  corn  is  to 
the  number  he  planted  with  potatoes,  as  |  to  1 ;  but  if  he 
had  planted  6  acres  less  of  corn,  and  J  as  many  potatoes 
4-15-J  acres,  the  ratio  would  have  been  as  f  to  f .  How 
many  acres  of  each  did  he  plant  ? 


[Note. — The  five  following  examples  are  designed  to  be  solved 
by  using  two  or  more  unknown  quantities.] 

11.  Find  two  numbers,  the  greater  of  which  shall  be  to  the 

less,  as  their  sum  to  42 ;  and  as  their  difference  is  to  6. 

Suggestion-  Let  x  =  one  and  y  the  other. 


Then 

(1)        x:y::x+y:^2 

(2)        x:y::x-y:6, 

By  equality  of  ratios 

x  +  y:4:2::  «— y:6, 

and 

x-\-y:x-y::  7:1, 

2x :  2y  : :  8  :  6, 

x:2y  ::  i:Q, 

a^:|y::4:4, 

.-.  ^  =  ty. 

Substituting  in  (2) 

iyy-'  |y-y:6 

Or, 

f:l::iy:6, 

4:1  ::    y:6 

1:1::    y :  24 

.-.  y  =  24, 

a?  =  |y  =  32. 

308  PBOPOBTION. 

12.  Two  numbers  have  such  a  relation  to  each  other,  that 
if  4  be  added  to  each,  they  will  be  in  proportion  as  3  to  4 ; 
and  if  4  be  subtracted  from  each,  they  will  be  to  each  other 
as  1  to  4.     What  are  the  numbers  ?  Ans.y  5  and  8. 

Suggestion,     x+4  :  y+4  : :  3 :  4  and  «— 4 :  2^—4  : :  1:4.   Whence 

4;c-|-4  4ic  +  4 

a?  f  1 :  3 : :  y :  4,  or  -— —  :l::y:l.     .-.  y  =  — ^- .      Substituting, 

aJ-4:  ^ 4::  1 : 4,  a;-4 :  2x-4  : :  1:6,  cc-4:a;::  1 :  5,  4:  x;:  4:  5. 

o 

.  •.  x  =  5.     Substituting,  6  :  3  : :  y :  4,  or  3  :  3 : :  y :  8.     .-.  y  =  S. 

13.  Find  two  numbers  in  the  ratio  of  2|-  to  2,  such  that 
when  each  is  diminished  by  5,  they  shall  be  in  the  ratio  of 
1^  to  1.  Numbers,  25  and  20. 

14.  There  are  two  numbers  which  are  to  each  other  as  16 
to  9,  and  24  is  a  mean  proportional  between  them.  What 
are  the  numbers  ?  Ans.,  32  and  18. 


[Note. — The  following  examples  may  be  solved  by  converting  the 
proportion  into  an  equation,  at  whatever  stage  of  the  solution  it  is 
found  expedient.] 

15.  Find  two  numbers  in  the  ratio  of  5  to  7,  to  which 
two  other  required  numbers  in  the  ratio  of  3  to  5  being 
respectively  added,  the  sums  shall  be  in  the  ratio  of  9  to  13-, 
and  the  difference  of  those  sums  =16. 

Numlers,  30  and  42,  and  6  and  10. 

16.  A  farmer  hires  a  farm  for  $245  per  annum  ;  the  arable 
land  being  valued  at  12  an  acre,  and  the  pasture  at  11.40  ; 
now  the  number  of  acres  of  arable  is  to  half  the  excess  of 
the  arable  above  the  pasture  as  28  :  9.  How  many  acres 
are  there  of  each  ? 

Ans.y  98  acres  of  arable,  and  35  of  pasture. 

17.  The  quantity  of  water  which  flows  from  an  orifice  is 
proportioned  to  the  area  of  the  orifice,  and  the  velocity 


APPLICATIONS.  309 

of  the  water.  Now  there  are  two  orifices  in  a  reservoir, 
the  areas  being  as  5  to  13,  and  the  velocities  as  8  to  7,  and 
from  one  there  issued  in  a  certain  time  561  cubic  feet  more 
than  from  the  other.  How  much  water  did  each  orifice 
discharge  in  this  time  ?      A7is.,  440  and  1001  cubic  feet. 

18.  At  an  election  for  two  members  of  parliament,  three 
men  offer  themselves  as  candidates,  and  all  the  electors  give 
single  votes.  The  number  of  voters  for  the  two  successful 
ones  are  in  the  ratio  of  9  to  8  ;  and  could  the  first  have  had 
seven  more  without  changing  the  numbers  which  the  others 
had,  his  majority  over  the  second  would  have  been  to  the 
majority  of  the  second  over  the  third  as  12  :  7.  Now  if  the 
first  and  third  had  formed  a  coalition,  and  had  one  more 
voter,  they  would  each  have  succeeded  by  a  majority  of  7. 
How  many  voted  for  each  ? 

Ans.y  369,  328,  and  300,  respectively. 

19.  A  man,  driving  a  flock  of  geese  and  turkeys  to  mar- 
ket, in  order  to  distinguish  his  own  from  any  he  might 
meet  on  the  road,  pulled  5  feathers  out  of  the  tail  of  each 
turkey,  and  2  out  of  the  tail  of  each  goose,  and  upon  count- 
ing them,  found  that  the  number  of  turkeys'  feathers 
lacked  15  of  being  twice  those  of  the  geese.  Having  bought 
20  geese  and  sold  15  turkeys  by  the  way,  he  found  that  the 
number  of  geese  was  to  the  number  of  turkeys  as  8  to  3. 
What  was  the  number  of  each  at  first  ? 

Ans.,  45  turkeys,  and  60  geese. 


PROBLEMS    OP    PURSUIT. 

20.  A  fox  starts  up  120  feet  ahead  of  a  hound  at  exactly 
J^  past  2  o'clock  P.  M. ;  the  hound  gives  chase  and  gains 
5  feet  every  2  minutes.  At  what  time  will  he  overtake  the 
iox? 


310  t>ROPORTIOJSr. 

statement.— Letting  x  be  the  time  wliich  will  elapse  before  the 
hound  overtakes  the  fox,  the  problem  becomes :  If  a  hound  gain  5 
feet  in  two  minutes,  how  long  will  it  take  him  to  gain  120  feet  ? 
That  is  5  :  120  : :  3  :  a;.  .-.  a;  =  48,  and  the  hound  overtakes  the  fox 
at  3  o'clock  and  18  minutes. 

21.  A  privateer  espies  a  merchantman  10  miles  to  lee- 
ward at  11.45  A.  M.,  and,  there  being  a  good  breeze,  bears 
down  upon  her  at  11  miles  per  hour,  while  the  merchant- 
man can  only  make  8  miles  per  hour  in  her  attempt  to 
escape.  After  2  hours  chase  the  topsail  of  the  privateer 
being  carried  away,  she  can  only  make  17  miles  while  the 
merchantman  makes  15.  At  what  time  will  the  privateer 
overtake  the  merchantman  ?  Ans.,  At  5.30  P.  M. 

22.  A  hare,  50  of  her  leaps  before  a  greyhound,  takes  4 
leaps  to  the  greyhound's  3 ;  but  2  of  the  greyhound's  leaps 
are  as  much  as  3  of  the  hare's.  How  many  leaps  must  the 
greyhound  take  to  overtake  the  hare  ? 

Suggestion.  Let  Sx  =  the  number  of  the  hound's  leaps, 

whence  4a;  =    "        "  "       hare's        " 

in  the  same  time.     Then  2  :^  ::  Sx:  4x  +  50.     .'.  x  =  100;  and  the 
hound  takes  300  leaps. 

23.  The  hour  and  minute  hands  of  a  clock  are  exactly 
together  at  12  M.     When  are  they  next  together  ? 

Suggestion. — Measuring  the  distance 
around  the  dial  by  the  hour  spaces,  the 
whole  distance  around  is  12  spaces. 
Now,  when  the  hour  hand  gets  to  1,  the 
minute  hand  has  gone  clear  around,  or  , 
over  12  spaces.  But  as  the  hour  hand  ' 
has  gone  one  space,  the  minute  hand  has 
gained  only  11  spaces.  Now  as  the 
minute  hand  must  gain  an  entire  round, 
or  12  spaces,  to  overtake  the  hour  hand, 
we  have  the  question:  If  the  minute  hand  gains  11  spaces  in  1  hour, 
how  long  will  it  take  to  gain  12  sj^aces?  .-.  11 :  12  : :  1  hour  :  x 
hours ;  and  x  =  1^\  hours,  or  1  hour  Oy\  minutes. 


APPLICATIONS.  Sll 

Scholium. — 1.  It  is  evident  that  the  hands  are  together  every  l^^ 
hours;  hence  to  find  at  what  time  thoy  are  together  between  any 
two  hours  on  the  dial,  we  have  only  to  multiply  1^*^  by  the  number 
of  the  whole  hours  past  12  o'clock.  Thus  between  7  and  8  they 
pass  each  other  at  7^^,  or  7  o'clock  and  3Sr^  minutes.  Between  10 
and  1 1  they  pass  each  other  at  10  o'clock  54^  minutes. 

24.  At  what  time  between  6  and  7  o'clock  is  the  minute 
hand  jiirit  J  of  the  circle  in  advance  of  the  hour  hand  ? 

Suggestion. — The  question  is :  If  the  minute  hand  gains  11  spaces 
in  one  hour,  how  long  will  it  take  it  to  gain  6|  rounds,  or  75  spaces? 
Or,  if  it  gains  1  round  in  -fij-  hours,  how  long  will  it  take  it  to  gain 
6|  rounds? 

Ans.,  At  49y'T^  minutes  past  6. 

25.  At  what  time  between  4  and  5  is  the  hour  hand  of  a 
watch  just  20  minutes  in  advance  of  the  minute  hand  ? 

Ans.,  At  no  time  hetiueen  these  hours.  The  minute  hand 
is  within  20  minutes  of  the  hour  hand  at  4  o'clock,  and  at 
5^  minutes  past  5. 

26.  Before  noon,  a  clock  which  is  too  fast,  and  points  to 
afternoon  time,  is  put  back  5  hours  and  40  minutes ;  and  it 
is  observed  that  the  time  before  shown  is  to  the  true  time 
as  29  to  105.     Required  the  true  time. 

Suggestion. — Letting  «=  the  time  pointed  to,  ar: :  a!  +  6| : :  29 :  105. 
Observe  that  to  tuni  the  hands  back  5h.  40m,  is  tlie  same  as  to  turn 
them  forward  6h.  20m.  x  =  2h.  25m.,  and  the  true  time  was  2h. 
25m.  +6h.  20m.,  or  15m.  before  nine  o'clock  in  the  morning. 

27.  Two  bodies  move  uniformly  around  the  circumfer- 
ence of  the  same  circle,  which  measures  s  feet.  When  they 
start,  one  is  a  feet  before  the  other  ;  but  the  first  moves  m 
and  the  second  M  feet  in  a  second.  When  will  these  bodies 
pass  each  other  the  1st  time,  when  the  2nd,  when  the  3rd, 
etc.,  supposing  that  they  do  not  disturb  each  other's  motion, 
and  go  around  the  same  way  ? 

Suggestion.— 1st.  If3/">w,  the  second  gains  M—m  feet  a  sec- 
ond, and  having  a  feet  to  gain,  overtakes  the  first,  and  does  it  in 


312  PROPORTION. 


— seconds.     The  problem  is  then  like  the  preceding,  as  the 

second  gains  a  whole  round  every  — seconds.    Hence  the  second 

°  M—m 

o  -^  s  n  -^  2ji 

passing  is  at  — from  the  starting,  the  third  at  -— — ,  the  fourth 

°  M—m  °  M—m 

.   a  +  ds      . 
at  -- —  ,  etc. 

M—m 

2nd.  If  Jf  <  7n,  the  second  is  overtaken 
by  the  first  after  the  first  has  gained  s—a 

feet,   or  in —  seconds ;  and  in  every 

seconds  thereafter ;  that  is,  from  the 

m—M 

..  c   ^     .-        '      2«-«     'da— a 

time  or  startmg,  in , ,  etc. 

m—M    m—M 

28.  The  earth  makes  a  revolution  around  the  sun  in  about 
365  days  and  Mars  in  about  687  days.  How  long  is  it  from 
the  time  at  which  these  planets  are  together  on  the  same 
side  of  the  sun  till  they  are  next  together?  That  is,  how 
long  does  it  take  the  earth  to  gain  an  entire  revolution? 

Ans.,  687—365  :  365  ::  687  :  x,  ,\  x  =  780  days,  nearly. 

[Note. — The  time  required  by  a  planet  to  go  around  the  sun  is 
called  its  Periodic  Time^  or  its  year.  When  two  planets  are  on  the 
same  side  of  the  sun  at  the  same  time,  they  are  said  to  be  in  Con- 
junction. The  time  from  one  conjunction  to  the  next  is  the  Synodic 
Period.  The  way  in  which  this  problem  usually  presents  itself  is 
this:  We  can  obserT^e  when  a  planet  is  in  conjunction  with  the  earth, 
and  thus  knowing  the  time  of  the  earth's  revolution  (a  year),  we  can 
find  the  Periodic  Time  of  the  planet,  or  how  long  it  takes  to  go 
around  the  sun.] 

29.  Calling  the  earth's  periodic  time  365^  days,  and 
observing  Jupiter's  synodic  period  to  be  about  399  days, 
how  long  is  Jupiter  in  completing  a  revolution  around  the 
sun ;  that  is,  what  is  its  periodic  time,  or  length  of  its  year? 

Ans.  ll^-J^-i^  of  our  3'ears. 

30.  Saturn's  synodic  period  is  about  378  days ;  what  is  its 
periodic  time  9  A71S.  29^  of  our  years. 


APPLICATIONS.  313 


PARTITIVE    PROPORTION. 

S2a,  Partitive  Proportion  is  a  term  applied  to  the 
division  of  a  number  into  parts  which  shall  be  in  the  ratio 
of  given  numbers. 

31.  Divide  350  into  two  parts  which  shall  be  to  each 
other  as  3  to  4.    As  2  :  5.    A^,  11  :  24. 

Suggestion. — Let  3a;  represent  one  part,  and  ix  the  other ;  whence 
dx-\-4x=  350,  X  =  50 ;  and  the  parts  are  150,  and  200. 

32.  Divide  a  into  two  parts  which  shall  be  to  each  other 
as  m  :  n. 

The  parts  are  — -, —  ,  and  '- — -  • 
^  m-{-n  m-\-n 

33.  Divide  560  into  three  parts  which  shall  be  to  each 
other  as  17,  23,  and  16.    As  43,  2,  and  11. 

34.  Divide  a  into  three  parts  which  shall  be  to  each 
other  as  m,  n,  and  r. 

35.  Divide  23  into  two  parts  which  shall  be  to  each 
other  as  f  and  f . 

The  parts  are  12^^  and  10|J. 

36.  Divide  a  into  two  parts  which  shall  be  to  each  other 

in  the  ratio  —  to  -•    As  —  :  -• 
m        n  m    n 

The  parts  in  the  last  case  are —  ,  and 


m8-\-m  ms-\-rn 


HAPTIII  II 

^Sectjon  f. 


PERCENTAGE. 

S3.  According  to  our  definition,  the  Equation,  of  which 
it  is  the  special  province  of  Algebra  to  treat,  is  the  grand 
instrument  for  investigating  the  relations  of  quantities. 
Now,  in  simple  Percentage,  there  are  four  quantities  to  be 
compared ;  viz.,  the  Base,  the  Bate  Ber  Cent.,  the  Bercent- 
age,  and  the  Amount,  and  the  problem  is.  To  discover  and 
express  in  equations  the  relations  between  these  four  quan- 
tities so  that  if  a  sufficient  number  of  them  are  given  the 
others  may  be  found. 

[Note.— For  Definitions  see  Practical  Arithmetic,  p.  225  et  seq.] 

84.  Prob.  1.— To  express  the  relation   between  base, 

pate  pep  cent.,  and  percentage. 

Solution. — Let  5  represent  the  base,  r  the  %,  and  p  the  percentage. 

If 
Now  rfo  means  r  lOOths  of  the  base.     Hence  r  %  of  5  is  -^  times  5, 

?•&  rh 

or  —  .     .'.  7)  =  —  • 
100  ^       100 

85.  Prob.  2.— To  exppess  the  pelation  between  pate 
pep  cent ,  amount  or  difference,  and  base. 

Solution. — Let  s  represent  the  sum  or  difference  of  the  base  and 


percentage.     Then  s  =  &  ± 


rh 
100 


¥l^^±ll ,  the  +  sign  to  be  used 
103       '  ^ 


PEKCKNTAGE.  316 

when  the  base  is  increased  by  the  percentage,  and  the  —  sign  when 
the  base  is  diminished  by  the  percentage. 

Scholium. — The  two  formuloi 

expressing  the  relation  between  the  four  quantities  J,  r,  /),  and  s, 
two  of  which  must  always  be  given  to  find  the  others,  are  in  then.- 
selves  suflBcient  for  the  solution  of  all  problems  in  Simple  Percentage. 

EXAMPLES. 

Ex.  1.   Bought  a  horse  for  $840,  and  sold  it  for  1560. 

How  much  did  I  lose  per  cent.  ?  Ans.,  33^%'. 

Suggestion. — Here  h  and  «  are  given  to  find  r.  Hence  formula  (2) 
is  to  be  used.     And  as  there  is  loss  involved,  the  —  sign  is  to  be 

taken.     Substituting  in  this  formula,  560  = .     Solving 

„  ,         56000       ,,^^  200       ,^^ 

for  r  we  have  -^vt^   =  100— r,  or  — -  =  100— r;    whence  r  =  100 
840  3 

_???  =  "»  =  33J. 
3  3^ 

2.  A  number  being  increased  by  2  equals  14.  Required 
the  increase  per  cent.  ?  Atis.,  16f^. 

rb 
Suggestion.— Formula  (1)  gives  2  =  — - ;     and    (2)     gives     14 

=  -— TTjA  -^  •     From  which  we  are  to  find  ;-.     Multiplying  (1)  by  7, 

-iA      '^rb       ^^  7;-&       &(100  +  r)        „        ,^^  100 

14  =  -  .     Whence  —  =  ^-^  or  7.  =  100  +  r,  or   r  =  — 

=  16|. 

3.  A  piece  of  cloth  sold  for  $7?9,  cash,  which  was  b%  off. 
Required  the  price  of  the  cloth.  Price,  $820. 

4.  Sold  40^  of  my  wheat,  and  had  remaining  981  bushels. 
How  much  had  I  at  first  ?  Arts.,  1635  bushels. 

5.  A  man  sold  two  horses  at  $420  each  ;  for  one  he  re- 
ceived 25^  more,  and  for  the  other  25;^  less  than  its  value. 
Required  his  loss.  Loss,  $56. 


316  BUSINESS    RULES. 


Suggestion. — Letting  b  and  h^  be  the  values  of  the  horses,  we  have 

._.       6(100-25)  ,  ...       &j(100  +  35)  ^^,       ,„^- 

420  = — - — ^ ,  and  420  =    ^^ .      .-.   756  =  125&i,  or   b 

K  ft 

= -&j  and  54-^1  =0^1,  the  value  of  the  horses.     Now  h^,  foupd 

O  o 

Q 

from  the  2nd,  is  336,  and  as  ~b^  —  840  =  the  loss,  we  have  896—840 
o 

=  56  =  the  loss. 

Those  who  are  not  familiar  with  algebraic  reasoning  will  prefer 
to  find  the  values  of  J  and  h^  from  the  two  formulas,  and  add  them 
together. 

6.  A  man  sold  72  turkeys,  which  was  32%  of  the  number 

he  had  remaining.     How  many  had  he  at  first  ? 

Ans.y  297. 

7.  A  farmer  saved  annually  $145^,  which  was  33^%  of 
his  annual  income.     Required  his  income  ?    Ans.y  $436^. 

8.  A  merchant  having  400  barrels  of  cider,  sold  at  one 
time  4:5%  of  it ;  at  another  time  20^  of  the  remainder.  How 
many  barrels  did  he  sell  in  all  ?  Jns.,  224  hbL 

rb       45x400       ... 
Operation.        p   ==TK?i=      ^^^     =  180. 

20x220  _ 

~ioo~-  **• 

•'•  p+Pt=224:. 

9.  A  housekeeper  gave  to  her  neighbor  -J  of  a  pound  of 
tea,  and  had  }  of  a  pound  remaining.  What  per  cent,  of 
her  tea  had  she  remaining?  Ans.,  85-Jf-^. 

«         ..  3        rX     ^        7r  600       „^, 

Operation.     4  =  1^0' '  =  lOO*    '•  "  =  ^  = ''^- 

10.  John  has  ^  of  a  dollar,  and  Henry  has  |  of  a  dollar. 
What  per  cent,  of  John's  money  equals  Henry's  ?  What  of 
Henry's  equals  John's  ?  Ans.,  120^,  83-J^. 

Operation.    l  =  f^^«  =  w     '■'■  =  ''°- 

1       ri     ^      Sr  250      .„, 

[Note. — For  other  examples  in  percentage,  see  Olney's  Practical 
Arithmetic] 


p 

100 

Pi 

100 

1I€TM>M    IL 


SIMPLE    INTEREST  AND   COMMON    DISCOUNT. 
[Note. — For  definitions  see  Practical  Arithmetic] 

86.  Prob.  1.  -To  express  the  relation  between  princi- 
pal, rate  per  cent.,  time,  and  interest. 

Solution.— Let />,  /■,  t,  and  i  represent  respectively  the  principal, 
rate  per  cent,  time  and  interest ;  t  being  in  the  denomination  for 
which  the  rate  per  cent,  is  estimated.  Thus,  if  the  rate  per  cent,  is 
rate  per  cent,  per  year,  t  is  to  be  underetood  as  years ;  if  the  rate 
per  cent,  is  per  month,  t  is  months,  etc. 

Then  as  p  is  the  base,  the  percentage  for  a  unit  of  time  is  -^  (§4)  ; 
and  for  t  units  of  time  it  is  —^ .     .-.  i  =  -^ . 

87.  Prob.  2.— To  express  the  relation  between  amount, 
principal,  rate  per  cent.,  and  time. 

Solution. — Since  the  amount  is  the  sum  of  principal  and  interest, 

representing  t^e  amount  by  o,  we  have  a=p-\-i.    But  i=  j^; 

100 

trp  100 +  tr  100a 

hence  «  =  />+-—,  or  a  =p-——.     r.  p  = 


100'  ^     100     *     "  ^      100 +  tr 

88,  Scholium- — This  problem  embraces  the  common  rule  for 
Discount  (see  Arithmetic,  383).  The  pupil  should  be  careful 
to  understand  the  reasonableness  of  discount.  For  example,  if  I  hold 
a  note  against  Mr.  B.,  which  note  is  payable  at  any  future  time, 
if  the  note  is  drawing  interest  for  all  money  is  worth,  tlie  Present  Worth 
of  the  note  at  the  time  it  was  given  was  its  face.  If  the  rate  at  which 
it  is  drawing  interest  is  less  than  money  is  worth  (or  if  it  draws  no 
interest)  the  Present  Worth  at  its  date  is  less  than  the  face  of  the 
note.  If  it  draws  7io  interest,  its  Present  Worth  at  any  time  is  less 
than  its  face.  Finally,  if  the  rate  which  the  note  draws  is  greater 
than  the  market  rate,  the  Present  Worth  of  the  note  at  its  date  is 
more  than  its  face. 

Scholium.— The  two  formtdft    (1)  i  =  r^  and 

100 +  tr 
i2)a  =  p  +  i=p-^-, 


318  BUSINESS    RULES. 

are  sufficient  to  solve  all  jjroblems  in  Simple  Interest  and  Common  Dis- 
count. 

EXAMPLES. 
Ex.  1.  What  is  the  interest  on  $250  for  1  yr.  10  mo.  15  da., 
at  Q%  per  annum.  Ans.,  $28.12|. 

Solution. — In  this  example  I  am  to  consider  principal,  time, 
rate  per  cent,  and  interest,  the  latter  of  which  is  the  unknown 
quantity.  Formula  (1)  expresses  the  relation  between  these  quanti- 
ties.    The  rate  per  cent,  being  pet-  anjium,  the  time  must  be  in  years. 

10  5 
15  days  =  |^  =  .5  of  a  month.     10.5  months  =  — ^  =.875  of  a  year. 

.'.    1   yr.    10  mo.    15   da.  =  1.875   years.     Now  (1)  gives  i  =zrJL 

a         6  10" 


_  1.875  x0x,.._ 

-  100  -  ^^•^^*- 

2.  What  is  the  interest  on  $47.25  for  1  yr.  and  6  mo.,  at 
%% per  annum?  Ans,,  $4.2525. 

.3  3  S^.4^ 

Operation,     i  = j^ =  4.2525* 

4 

3.  What  is  the  interest  on  $145.50  for  1  yr.  9  mo.  24  da., 
at  6% per  annum?  Ans.,  $15.86  nearly. 

.      1.816x6x145.50      _  ^^ 
Operation,    t  = .^ =  15.86  nearly. 

4.  What  is  the  interest  on  $123.75  for  2  yr.  8  mo.  12  da., 
at  6%  per  annum  ?  Ans.,  $20.0475. 

6.  W^hat  is  the  interest  on  $475  for  2  yr.  7  mo.  and  20  da., 
at  Q%  per  annum  ?  Ans.,  $75.208 J. 

6.  What  is  the  interest  on  $340.60  for  4  yr.  and  5  mo.,  at 
6^  per  annum  ?  -  Ans.,  $90,259. 

7.  What  is  the  interest  on  $50.40  for  1  yr.  and  10  mo.,  at 
"1%  per  annum?  Am.,  $6,468, 


SIMPLE   INTEREST   AND   Ct)MMON    DISCOUNT.  319 

Suggestion. — In  solving  this  example  the  operation  upon  paper 
consists  simply  in  multiplying  12.833  by  .504.  The  pupil  should 
always  reduce  the  written  details  of  his  arithmetical  work  to 
the  minimum.     Thus,  in  this  case,  he  sees  mentally  that  10  mo. 

are   .833   of  a  year.     Hence  i  =  ^^J^^^^^""-! ,    But   1.833 

X  7   he  produces  mentally,   and  writes   12.833.     And  12.833 

cance'lling  the   100  from   50.40   makes  it  .504.      Hence  .504 

the  only   written  work   should   be  as  in   the  margin.  51332 

In  practice,  nothing  but  this  multiplication  should  be  64165 

written  down.  6^467832 

Solve  the  following  by  thus  reducing  the  written  work  to  a 
minimum.    The  answers  are  given  as  in  practice  in  business. 

8.  What  is  the  interest  on  $49.80  for  2  yr.  and  11  mo.,  at 
1%  per  annum  f  Ans.^  $10.17. 

operation,    i  =  1?:!^^£H  =  2.9166  x  7  x  .498  =  10.17. 

9.  What  is  the  interest  on  195.40  for  3  yr.  9  mo.,  at  S% 
per  annum  ? 

.      95.40x8x3.75      ^_  .„     _,  ^        v,     ,^ 

Operation,     i  = Ym\ =  28  62.    The  operation  should 

be  performed  mentally.  Thus  8  x  3.75  =  30.  Dropping  the  0,  and 
one  0  from  the  denominator,  and  for  the  10  remaining  in  the  denomi- 
nator removing  the  decimal  point  in  95.40  so  as  to  make  it  9.54,  we 
have  simply  to  multiply  9.54  by  3.  All  of  this  should  be  done  at  a 
glance,  without  writing  more  than  is  given  above. 

10.  What  is  the  interest  on  $196  for  5  yr.  7  mo.,  at  9^ 
per  annum  9  Arts.,  $98.49. 

11.  What  is  the  interest  on  $471.11  for  4  yr.  8  mo.,  at  1^% 
per  annum  f  Ans.,  $164.89. 

12.  What  is  the  interest  on  $18.60  for  3  mo.  12  da.,  at  3^ 
per  mo.?  Ans.,  $1.90. 

3.4  X  3  X  18.60    ,  ^    „     „^^     «..«    ^  ,     .  «« 
Operation,     i  = ^ =1.7  x  3  x  .372=.872  x  6.1  =  1.90. 

13.  What  is  the  interest  on  $400  for  $150  days,  at  %i% 
per  month  f  Ans.,  $50. 


320  BUSINESS    EULES. 


Operation.     i  = j^^ =  50 


14.  What  is  the  interest  on  $1,000  for  2  tyw.  12  da.,  at 
l\% per  month?  Ans.,  $36. 

15.  What  is  the  amount  of  $432.10  for  5  yr.  4  mo.  24  da., 
at  1%  per  annum  9  Ans.,  $595.43. 

Operation,     a  =  p^-^  =  432.10  '-^^  =  4.321  x  137.8 
=  595.4338. 

16.  What  is  the  amount  of  $325.25  for  2  yr.  9  rno.  12  da., 
at  6i%  per  annum  ?  Ans.,  $384.09. 

„«~  .►  100  +  2.784x6.5      „^,  ^^ 
Operation,    a  =  325.25  — -^-^ =  384.09. 


17.  In  what  time  will  $13,  at  6%  per  annum,  give  $0,975 
interest  ?  Ans.,  1  yr.  3  mo. 

Solution. — Since  principal,  rate  per  cent.,  time  and  interest  are 

compared  i  =  -r^  gives  the  relation.     As  time  is  required,  I  solve 

this  equation  for  ^,  and  have  t  = .  Substituting  the  given  values, 

.025 
6  0  .BkT«. 

18.  In  what  time  will  $45.25,  at  6%  per  annum,  give 
$1.81  interest  ?  Ans.,  8  mo. 

19.  In  what  time  will  $70.50,  at  9^  per  annum,  give 
$31,725  interest?  Ans.,  5  yr. 

20.  In   what  time  will   $140,  at   7^  per  annum,,  give 
$10.861|  interest?  Ans.,  1  yr.  1  mo.  9  da. 


SIMPLE  INTEREST   AND   COMMON    DISCOUNT.  321 

21.  In  what  time  will  $48.50,  at  ^%  per  annnmy  amount 
to  $56.187i  ?  Ans.y  2  yr,  7  mo.  21  da. 

Suggestion. — Subtract  the  principal  from  the  amount  to  find  the 
interest. 

22.  In  what  time  will  $248,  at  Q%  per  anuuni,  amount  to 
$282,224  ?  Ans.,  2  yr,  3  mo.  IS  da. 

23.  In  what  time  will  $700,  at  d%per  ammm,  amount  to 
$712.35?  Am.,  2  mo.  10.5+  da. 


24.  At  what  per  cent,  will  $325  produce  $3.25  interest  in 
2  months?  Arts.,  6%. 

iff) 
Solution. — Same  as  above,  finding  r  from  the  equation  i  —  — J-, 

_  -lOOi  _  100  X  3.25  _ 
~    <;>    ~"    |x325      ~ 

25.  At  what  per  cent,  will  $40  produce  $13.36  interest 
in  2  yr.  9  mo.  12  da.  ?  Ans.,  \%%. 

26.  At  what  per  cent,  will  $125  produce  $32,375  interest 
in  3  yr.  6  mo.  ?  Arts.,  7|^. 

27.  At  what  per  cent,  will  $124  produce  $29. 17^  interest 
in  4  yr.  3  mo.  10  da.  ?  Ans.,  h\%. 

28.  At  what  per  cent,  will  $2,360.25  amount  to  $2,470,395 
iu  7  months  ?  Ans.^  %%. 

Suggestion.— Find  the  interest  and  then  proceed  as  before. 

29.  At  what  per  cent,  will  $230  amount  to  $249.83f  in 
11  mo.  15  da.  ?  Ans.,  9%. 


30.    What  princi}>al  will  in  3  yr.  8  mo.  15  da.,  at  6%  per 
annitm,  give  $76,095  interest  ?  A  ns.,  $342. 


BUSINESS    RULES. 

Solution.— Solving  i  =  -^  for^,  I  have 

_  im  _  100  X  76.095  _  100  x  76.095  _ 
^  ~  IT  ~~     6  X  3.7tV     ~        23.25        ~ 

31.  What  principal  will  in  4  yr.  9  mo.  18  da.,  at  9^  jpgr 
annum,  give  165.016  interest?  ^ws.,  $150.50. 

32.  What  principal  will  in  8  yr.  8  mo.  12  da.,  at  5%  /?er 
amium,  give  $147.9435  interest.^  A7is.,  $340.10. 


33.  How  long  will  it  take  for  $200,  at  simple  interest,  at 
Q%  per  annum,  to  amount  to  $500  ?  Ann.,  25  years. 

Solution. — I  have  under  consideration^,  r,  a,  and  t,  the  latter  of 
which  is  the  unknown  quantity.     The  relation  between  these  is 

a=  p  — T?m~  •     Solving  this  for  t,  I  find 

^  _  100(<?-j9)  _  100(500-200)  _  300  _ 
~        pr        ~        6  X  200        ~  12  "" 

34.  How  long  will  it  take  $1,  at  simple  interest,  at  10^ 
per  annum,  to  amount  to  $100  ? 

35.  How  long  will  it  take  $75,  at  interest  at  b%  per  an- 
num,  to  amount  to  $100  ? 

80,  Cor.  —  To  find  the  time  required  for  a  principal  to 
double,  triple,  or  become  n  times  itself  at  any  rate  of  simple  inter- 
est, we  have  a  =  2p,  3p,  or  np.  Hence  the  last  formula  becomes 

i  = ^ — —  = for  the  time  required  to  double.     To 

pr  V  ■ 

*  '  1         I,       ^       100  X  2      ,          ^      ,     ,       iOO  X  3      , 
triple  we  have  t  == ;    to  quadruplSf  t  = /    to 

become  n  times  itself  t 

r 

Scholium. — It  appears  that  the  time  in  such  a  case  is  independ- 
ent of  the  principal,  as  might  have  been  anticipated. 

36.  How  long  will  it  take  $100  to  become  $500  at  S% 
per  annurnf  A^s.,  50  years.. 


SIMPLE   INTEREST,    AND   COMMON   DISCOUNT.  323 

37.  How  long  does  it  take  a  principal  to  double  at  6%  per 
annum  ;  at  1%  ;  at  5^  ? 

38.  Sold  property  amounting  to  $3,000,  on  a  credit  of  12 
mo.  without  interest.  Money  being  worth  S%  per  annum, 
what  sum  in  hand  is  equivalent  to  the  $3,000  under  the 
contract?    Verify  it.  Ans.,  12,777.??  +  . 

Solution. — I  am  to  find  a  sum  which  put  at  interest  for  1  yr.  at 
8%  will  amount  to  $3,000.    I  therelbre  have  given  a,  ?•,  and  t  to 

find  p.    The  relation  of  these  is  given  in  the  formula  a  =  p 

^  ^.  ^  100a         ,.      -.  ^300000       «„„„„„ 

from  which  p  =  — -r — -  =  (in  this  case)  -—7,7;-  =  2777.77  +  . 
^       100 +  ^r      ^  ^108 


100 


39.  A  debt  of  $500  will  be  due  in  3  ym.  without  interest. 
What  is  its  present  worth  if  money  commands  6%  per  annum  ? 

Ans.,  $423.73. 

40.  What  discount  should  be  allowed  for  the  present  pay- 
ment of  a  note  of  $400,  due  3  yrs.  5  mo.  hence,  the  note  not 
bearing  interest,  though  money  is  woi-th  6^  per  annum  9 

Ans.,  $68.05. 

41.  A  man  buys  a  piece  of  property  to-day  for  $5,000, 
giving  his  note  with  security  at  12^^  per  annum,  payable 
2  yrs.  9  mo.  hence.  What  is  the  present  worth  of  this  note 
if  money  brings  in  market  6%  per  annum  ? 

Ans.,  $5,708.13—. 

Operation,    a  =i) —J-^'  =  50(100 +  12  x||) =50x138=6650. 

100a         665000       ^^^o^K 

Pt  =  z-;z7: =  -TTTTzr  —  5708.15—. 

^*        100 +  tr,         116.5 

42.  A  merchant  buys  $2,000  worth  of  goods  on  90  days 
time,  at  2%  a  month.  He  could  have  borrowed  money  at 
1^%  a  month.  How  many  more  goods  could  he  have  bought 
for  the  same  money  if  he  bad  borrowed  ? 

Ans.,  $28.71  worth. 


324  BUSIISTESS  RULES. 

43.  July  20tli,  1869,  I  hold  a  note  for  1500  dated  April 
6th,  1867,  due  Jan.  1st,  1872,,  and  bearing  interest  at  1% 
per  annum.  What  is  the  present  worth  of  this  note,  money 
being  worth  10^  per  annum  9  Ans. ,  $534.86  + . 


00,  Prob. — To  find  what  each  payment  must  be  in  order 
to  discharge  a  given  principal  and  interest  in  a  given  num- 
ber of  equal  payments  at  equal  intervals  of  time. 

Solution. — Let  p  represent  the  principal,  r  the  rate  percent., 
t  one  of  the  equal  intervals  of  time,  n  the  number  of  payments,  (i.  e., 
nt  is  the  whole  time),  and  x  one  of  the  payments. 

There  will  be  as  many  solutions  as  there  are  different  methods  of 
computing  interest  on  notes  upon  which  partial  payments  have 
been  made. 

1st.  By  the    United  States   Court  Rule. — As  the  payments  must 

exceed  the  interest  in  order  to  discharge  the  principal,  this  rule 

requires  that  we  find  the  amount  of  ^,  for  time  t,  at  r  per  cent. 

rt  I        ri  \ 

This  is  done  by  multiplying  by  l  +  fT^T^,  and  gives  i^ll  +  Tn?;)- 

From  this  subtracting    the  payment    x,   the    new    principal    is 

pll  +  -TT^ )  ~^-   Again  finding  the  amount  of  this  for  another  period 

of  time,  t,  and  subtracting  the  second  payment 

In  like  manner,  after  the  third  payment  there  remains 
After  the  4th  payment,  the  remainder  is 
Finally,  after  the  nth  payment,  we  have 

H'+ioo)-K'+ioo)    -T+ioo)    - 


SIMPLE   INTEREST,— PARTIAL  PAYMENTS.  325 

Whence 


Mo)" 


1  + 

This  denominator  being  the  sum  of  a  geometrical  progression  whose 
first  term  is  1,  ratio  (l  +  f7fr^)j  and  number  of  terms  n,  its  sum  is 
/.      rt  Y    ,  prtL      rt  \" 

0  +  ioo  -'     „  TobO+wo 

^ .    Hence  x  =  - — ^— -- — -  • 

100  \  ^100/ 

2nd.  By  the  Vermont  Rule. — Tlie  amount  of  the  principal  for  the 

whole  time  is  7)  1 1 4-  r^  y 

The  amount  of  the  1st  payment  is x\  ^  +  t7^('^—^)  \ 

"        "      2nd        "  aj[l  +  -^(n_2)], 

"        "      3rd        "  -  .  ^[l  +  -^(«_3)], 

Etc.,  etc.,  etc. 

The  nth  payment  (with  no  interest)  is  - x. 

The  sum  of  the  amounts  of  these  payments  is 

n^'^~Qx[(ji-l)^{n-%)-\-{n-^) 1]. 

The  series  in  the  brackets  being  an   arithmetical  progression 
whose  first  term  is  (n— 1),  common  difference  —1.  last  term  1,  and 

number  of  terms  (n— 1),  its  sum  is  (— p-  \n.     Hence  the  sum  of  the 

^       nrt 

payments  is  ^^  +  7aa^I  "«- ) '^j  o"*  x^^n-\ J.    But  by  the 

condition  this  sum  equals  the  amount  of  the  principal ;  conse- 
quently 


nrt 

X 


4-S 


x  = 

a        ^rt  .         . . 


336  BUSINESS   RULES. 

Scholium.— If  the  payments  are  made  annually,  i  =  1.  And  let- 
ting / '  =  '-- ,  /.  €.,  letting  the  rate  per  cent,  be  expressed  decimal- 
ly, the  formulas  become 

By  the  U.  8.  Rule         x  =  ^f^-±^ ; 

By  the  Vermont  Rule    x  =  —^ — -—4^  • 

44.  What  must  be  the  annual  payment  in  order  to  dis- 
charge a  note  of  15,000,  bearing  interest  at  10^  per  annum, 
in  5  equal  payments  ? 

Ans,,  By  the  U.  S.  Rule,  $1,318.99  within  a  half  cent. 
By  the  Vermont  Rule,  $1,250. 

Query. — What  occasions  the  great  disparity  between  the  pay- 
ments required  by  the  different  rules  ? 

45.  What  annual  payment  is  required  to  discharge  a  note 
of  $300,  bearing  interest  at  7  per  cent,  per  annum,  in  4  equal 
payments  ? 

46.  The  sum  of  $200  is  to  be  applied  in  part  towards  the 
payment  of  a  debt  of  $300,  and  in  part  to  paying  the  inter- 
est, at  6^  in  advance,  for  12  months,  on  the  remainder  of 
the  debt.  What  is  the  amount  of  the  payment  that  can  be 
made  on  the  debt  ? 

Suggestion. — Let  aj  represent  the  payment;  then  (300— ;»)  x  j|(y 
is  the  interest  on  the  remainder  of  the  debt ;  and  we  have  therefore 
the  equation,  a;-f-(300— a;)  x^-^  =  200. 

Ans.,  $193.62, 

47.  A  is  indebted  to  B  $1,000,  and  is  able  to  raise  but  $600, 
With  this  sum  A  proposes  to  pay  a  part  of  the  debt,  and  the 
interest,  at  %%  iii  advance,  on  his  note  at  2  years  for  the 
remainder.     For  what  sum  should  the  note  be  drawn  ? 

Ans,,  $476.19, 


'^rineiysnip 


[Note.— For  definitions,  see  Practical  Arithmetic] 

iH,  Principle. — ///  Simple  Partnership,  i.  e.,  when 
all  the  capital  is  em  ployed  the  same  length  of  time, 
the  fandamental  principle  is  that  the  shares  of  the 
gains  shall  hear  the  sa^me  ratio  as  the  shares  of  the 
stock. 

EXAMPLES. 
Ex.  1.  A  and  B  enter  into  partnership.     A  puts  in  11,200 
and  B  $1,800.     They  gain  !&900.     What  is  each  one's  share 
of  the  gain  ? 

Solution. — Let  x  and  x^  be  their  respective  sliares  of  the  gain. 
Then  aj+a;,  =  900 

and  x'.x^  ::  1200:  1800  : :  2  :  3. 

Whence  x-^x^  :  x  : :  5  :  2 
or  900  :  X  : :  5  :  2 

.-.     x  =  360,  and  aj,  =  540. 

2.  A,  B,  and  C  enter  into  partnership.  A  puts  in  $340, 
B  $460,  and  C  $500.  They  gain  $390.  What  is  the  gain 
of  each?  Arts.,  A's  $102,  B's  $138,  and  C's  $150 

Operation.  x-\-x^  +«2  =  ^^-  xix^ix^::  340  :  4G0 :  500,  or 
ar+x^+Xg:  a?::  1300:340,  or  890  :  ar : :  180  :  34,  or  3  :  x  : :  1 :  34. 
.-.  a?  =  102.  a;+ar,+Xj:  a?,  ::  1300:460,  or  390  :  ar,  : :  130  :  46. 
or  8:a;j::l:46.  .-.  aj^  =  138.  a;-f  ar^ +a;g  :  a-g  : :  1300  :  500,  or 
390  :  ajg  : :  130  :  50,  or  3  :  ajjj  : :  1 :  50.     .  •.  x.^  =  150. 

3.  Any  number  of  persons  as  A,  B,  C,  D,  etc.,  unite  in  a 
partnership,  putting  in  respectively  a,  b,  c,  d,  etc.,  dollars 
each.     The  gain  or  loss  is  s.     What  portion  falls  to  each  ? 


328  BUSINESS   RULES. 

Solution.— Let  Xa ,  «i ,  ^c ,  -^d ,  etc.,  be  the  respective  shares  of  the 
gain  or  loss.    Then  «„ + ^j^  +  jc^ + a^^  + ,  etc.  =  «, 

and    Xa:Xb:Xc:Xd-\- etc.  \:  a:l.c:d etc. 

Whence  a;a+«6+aJc  +  «d  +  etc. :»«  ::  ti  +  ft  +  c+^+etc.  :a 
or       s:  iCa  ::  «  +  6  +  c  +  6Z4-,  etc.  :a. 

_  ^^ 

*  ~a+6+c+<Z  +  ,  etc. 

In  like  manner       x,,  —  — 7 3 :— , 

a+o  +  c+a+,  etc. 


a+6  +  c+(Z+,  etc.  * 
_  ds 

''~  a  +  h  +  c  +  d+j  etc. 

Scholium. — This  is  the  common  rule  for  Simple  Partnership,  viz.: 
Multiply  the  gain  or  loss  by  each  partner's  stock  and  divide  the 
products  by  the  whole  stock. 

[Note. — In  such  examples  the  solution  is  so  simple  that  the 
equation  scarcely  renders  any  assistance.  In  the  following  its 
advantages  will  appear.] 

4.  Two  men  commenced  trade  together.  The  first  put  in 
140  more  than  the  second  ;  and  the  stock  of  the  first  was  to 
that  of  the  second  as  5  to  4.     What  was  the  stock  of  each  ? 

Ans.,  $200,  and  $160. 

Operation. — Let  x  =  what  the  first  put  in.  Then  x—4,0  =  what 
the  second  put  in.  And  we  have  x :  a;— 40  : :  5  :  4,  or  « :  40  : :  5  :  1. 
.-.  a?  =  200. 

5.  Three  men  trading  in  company  gained  1780,  which  was 
to  be  divided  in  proportion  to  their  stock.  A's  stock  was 
to  B's  as  2  to  3,  and  A's  to  C's  as  2  to  5.  What  part  of  the 
gain  should  each  have  received  ? 

A71S.,  A,  $156  ;  B,  $234  ;  C,  $390. 

6.  Three  men  trading  in  company,  put  in  money  in  the 

following  proportion  :  the  first,  3  dollars  as  often  as  the 

second  7,  and  the  third  5.     They  gain  $960.     What  is  each 

man's  share  of  the  gain  ? 

Ans.,   $192;  $448;  $320. 


PARTNERSHIP.  ^ 

7.  A,  B,  and  C  found  a  purse  of  money  ;  and  it  was  mu- 
tually agreed  that  A  should  receive  115  less  than  one  half, 
that  B  should  have  $13  more  than  one  quarter,  and  that  C 
should  have  the  remainder,  which  was  $27.  How  many 
dollars  did  the  purse  contain  ?  A7is.j  1100. 


02,  Principle. — In  Coinpound  Partnership,  i.  e., 
partnership  in  jvhich  the  several  partners'  shares  of 
the  capital  are  in  for  different  lengths  of  time,  the 
gain  or  loss  is  divided  in  the  ratio  of  the  products  of 
the  several  amounts  of  stock  into  the  time  which  they 
respectively  remained  in  the  business.  TJiis  is  assum- 
ing that  the  use  of  la  for  time  t  in  business  is  equal  to 
lat  for  time  1. 

8.  A  and  B  enter  into  partnership.  A  furnishes  1240  for 
8  months,  and  B  $560  for  5  months.  They  lose  $118.  How 
much  does  each  man  lose?  Ans.,  A  48,  and  B  $70. 

Solution. — Let  iCa  =  A's  share  of  the  loss,  and  x^^  B's.  Then 
a?.  +  a?i  =  118,  and  «« : «« : :  8  x  240 :  5  x  560  : :  24 :  35.  Whence 
x«+a-6:aj«  ::  59:24,orll8::r:„  r:  59:24,  or2:a!„  ::  2:48.  .-.  «„  =  48, 
and  Xft  =  118-48  r=  70. 

9.  A,  B,  and  C  entered  into  partnership.  A  put  in  $100 
for  4  months,  B  $300  for  2  months,  and  C  $500  for  3  months. 
They  gained  $250.     How  much  was  each  man's  gain  ? 

Operation.    aj«+aJi4-«e  =  250,  and 
x^-.Xi'.x^  ::4xl00:2x300:3x500  ::  4:6;15.     Whence 
250  :ar^  : :  25  : 4,    or  10  : ««  : :  1  : 4.     .-.  Xa  =-.  40. 
250:ajft  ::25:6,    or  lOiajj  : :  1 :6.    .-.  spj  =  60. 
250  :«,  : :  25  :  15,  or  10  :  x,  : :  1 :  15.  .-.  x,  =  150. 

IQ.  A,  B,  and  C  hire  a  pasture  for  $180.  A  puts  in  8 
cows  for  10  weeks,  B  20  for  5  weeks,  and  C  30  for  9  weeks. 
How  much  ought  each  to  pay  ? 

Ans.y  A  $32,  B  $40,  and  C  $108. 


330 


BXrsil^ESS  RtJLBS. 


11.  To  gather  a  field  of  wheat,  A  furnished  8  laborers  for 
5  days,  B  12  for  3  days,  and  C  6  for  4  days.  For  the  whole 
work  A,  B,  and  C  received  $45.50.  How  much  should  each 
have  received  ?    Ans.,  A  $18.20,  B  $10.38,  and  0  $10.92. 

12.  Any  number  of  persons,  as  A,  B,  C,  D,  etc.,  unite  in 
partnership,  A  putting  in  $a  for  time  t^ ;  B,  $^  for  time  t^ ; 
C,  $c  for  time  t, ;  D,  M  for  time  td ,  etc.  The  gain  or  loss 
is  s.     How  is  it  to  be  shared  by  the  partners  ? 

Solution. — Letting  oj^,  a^  x„  Xd,  etc.,  represent  the  respective  shares 
of  the  gain  or  loss,  we  have 

^a  +  ^b+^c  +  ^d  +  ,  etc.,  —  s,  and 
Xa  •.Xk'.Xa'.Xd ctc,  : :  ata :  Uf, : etc  '-did etc.     Whence 

s:Xa::  at^  +  Uf,  +  ct,  +  dtd--  etc. , : at^.    .:  x^  =  — — -^ —  g. 

'  aia  +  lk-¥ct^-¥  did --etc. 

Ik 
^. 

ain + 0% + cU  +  dtd--  etc. 


siXf,::  ata  +  U(,  +  ct,  +  dtd-- etc. : Uj,. 
And  in  like  manner  for  the  others. 


.'.  ajft  = 


Scholium.— This  is  the  common  rule  for  Compound  Partnership, 
viz. :  Take  the  product  of  each  partner's  share  of  the  stock  into  its 
time  in  trade.  For  any  partner's  share  of  the  gain  or  loss  multiply 
the  whole  gain  or  loss  by  the  product  of  his  stock  into  its  time,  and 
divide  by  the  sum  of  the  several  shares  of  the  stock  into  their 
respective  times. 


Ex.  1.  A  farmer  mixes  together  10  bushels  of  oats,  at  40 
cents  a  bushel,  15  bushels  of  corn  at  50  cents  a  bushel,  and 
25  bushels  of  rye  at  70  cents  a  bushel  ?  What  is  the  value 
of  a  bushel  of  the  mixture  ?  Ans.,  58  cents. 


A.LLTGATiON;  331 

Solutioh. — Let  :t  be  the  value  ola  bushel  of  tlie  mixture  of  which 
there  are  10  + 15  +  25  —  50  bushels.  Hence  SOar  represents  the  value 
of  the  whole  mixture,  and  50.r  =  10  x  40  +  15  x  50  ^'  25  x  70  =  2900. 
.-.  X  =  58. 

2.  A  grocer  mixes  120  pounds  of  sugar  at  5  cents  a 
pound,  150  pounds  at  6  cents,  and  130  pounds  at  10  cents. 
What  is  the  value  of  a  pound  of  the  mixture  ?  Atis.,  $0.07. 

3.  A  liquor  dealer  mixes  8  gallons  of  alcohol  100^,  12 
.'gallons  80^,  25  gallons  60^,  40  gallons  40^,  and  60  gallons 
20%  strong.     What  is  the  strength  of  the  mixture  ? 

Am.,  41|4^. 

4.  One  kind  of  wine  is  40  cents  a  quart,  and  another  24. 
How  much  of  each  must  be  taken  to  make  a  quart  worth 
28  cents  ? 

Statement,    x+y  -1.     40a;  +  24y  =  28. 


5.  Three  kinds  of  sugar  are  worth  respectively  6,  8,  and 
10  cents  a  pound.  How  much  must  be  taken  of  each  to 
make  a  mixture  worth  7  cents  a  pound  ? 

Solution. — Let  cc,  y,  and  z  be  the  amounts  of  each  required. 
X  pounds  at  6  cents  a  pound  arc  6y;  y  at  8,  8y ;  z  at  10,  lOs.  The 
whole  amount  is  x  +  y-f-sr,  and  the  whole  value  6.1  +  82/+ 10^.  Hence 
6a;  +  8y  +  102  =:  7(rr  +  y  +  2).  But  here  are  <A?-<?e  unknown  quantities, 
and  the  example  gives  but  one  set  of  conditicms.  The  problem  is 
therefore  indeterminate ;  that  is,  there  are  not  conditions  enough 
given  to^  the  values  of  the  unknown  quantities. 

Suppose  we  add  the  two  conditions :  To  make  a  mixture  of 
48  lbs. ;  and  that  twice  as  much  of  the  6  cent  su^ar  shall  be  used 
as  of  both  the  others.  We  then  have  the  two  additional  equations 
aj  +  y  +  2  =  48,  and  x  =  2(y  +  z).  These  equations,  together  with  the 
former,  Q.r -\-Qy  +  lOz  =  7(x  +  y  +  z)  readily  give  x  =  82,  y  =  8,  and 
e  =  8. 

03,  Scholium.— The  last  example  is  a  case  in  Alligation  Alter- 
nate, as  it  is  denominated  in  our  Arithmetics.  Such  examples,  as 
they  are  "usually  stated,  are  simply  problems  in  which  there  are 
more  unknown  quantities  than  equations,  and  are  hence  IndeUrmU 


i^SH  BITSIKESS  RULES. 

7iate.  The  Indeterminate  Analysis  can  not  be  treated  in  this  volume, 
but  a  few  further  illustrations  of  such  examples  will  be  given.  All  the 
various  methods  of  treating  Alligation  Alternate  usually  given  in  our 
Arithmetics  are  exceedingly  cumbrous  and  perplexing  to  the  pupil, 
and,  after  all,  fail  to  give  a  full  view  of  it.  The  propriety  of  puz- 
zling pupils  with  any  of  them  is  exceedingly  questionable.  They  are 
very  clumsy  and  incomplete  efforts  at  doing  a  thing  which  becomes 
very  simple  when  the  proper  principles  are  developed,  which  prin- 
ciples cannot  be  brought  forward  in  Common  Arithmetic. 

6.  How  mucli  of  each  sort  of  grain,  at  48,  50,  and  68  cents 
a  bushel,  must  be  mixed  together,  so  that  the  compound  will 
be  worth  60  cents  a  bushel  ? 

Statement,  x,  y,  and  z  being  the  amounts  of  each  kind  respec- 
tively, we  have  48x  +  50y  +  QSs  =  60(x  +  p  +  z),  or  Qx  +  5y  =  4:Z.  Now 
any  real,  positive  values  may  be  assigned  to  either  two  of  these 
unknown  quantities,  which  will  give  a  positive  value  for  the  other. 

Thus  if  2  =  4,  and  y  =  2,  Qx=  16  —  10,  or  x  =  l.  Verify  these 
results. 

Again,  if  2  =  5,  and  y  =  t,  6x  =  20  —5,  or  a;  =  2|.  Verify  these 
results. 

Again,  if  «  =  6,  and  y  =  d,  Qx  =  24—15,  or  x  —  1^.  Verify  these 
results. 

In  like  manner  an  unlimited  number  of  sets  of  answers  can  be 
obtained. 

If,  however,  we  try  2  =  2,  and  y  =  3,  we  have  x=  —1}.  The 
negative  sign  in  this  case  shows  an  impossibility.  The  cause  of  this 
is  evident  when  we  notice  that  2  bushels,  at  68  cents,  and  3  at  50, 
make  5  bushels,  worth  286  cents,  or  57|  cents  per  bushel.  This 
mixture  cannot  be  made  worth  60  cents  per  bushel  by  putting  in 
grain  worth  only  48  cents. 

The  — 1|  of  the  48  cent  grain  indicated  by  this  result,  may  be 
understood  to  mean  that  we  are  to  take  out  of  the  5  bushels,  worth 
286  cents.  1|  bushel,  worth  48  cents  per  bushel.  This  leaves 
3|  bushels,  worth  230  cents,  or  just  60  cents  per  bushel. 

7.  A  merchant  has  two  kinds  of  wine.  The  first  kind  is 
worth  12  shillings  per  gallon,  and  the  second  is  worth  7 
shillings  per  gallon.  How  many  gallons  of  each  kind  must 
he  use  in  order  to  form  a  mixture  worth  ^  shillings  per 
gallon  ? 


ALLIGATION.  333 

Ans.j  Letting  x  and  i/  represent  the  quantities  respec- 
tively, he  may  take  any  quantity  he  pleases  of  either,  so  that 
he  takes  such  an  amount  of  the  other  as  to  preserve  the 
relation  '3x  =  2y.  That  is,  he  must  take  IJ  times  as  much 
of  the  latter  as  of  the  former.  If  he  takes  2  of  the  former, 
he  must  take  3  of  the  latter.  If  he  takes  6  of  the  former, 
he  must  take  9  of  the  latter,  etc. 

8.  How  much  corn  at  48  cents,  barley  at  3U  cents,  and 
oats  at  24  cents  per  bushel,  must  be  taken  to  make  a  com- 
pound worth  30  cents  per  bushel  ? 

Suggestion. — Show  that  he  can  take  any  amount  he  pleases  of 
either  one,  if  he  takes  proper  amounts  of  the  other  two  respectively. 
Show  what  he  may  take  of  the  second  and  third  kinds  if  he  take 
1  bushel  of  the  first.  Is  there  any  limit  to  the  number  of  ways  he  can 
make  up  the  mixture  when  he  takes  1  bushel  of  the  first  kind  ?  Can 
he  take  3  bushels  of  the  first  kind  and  2  of  the  second  ^  What  value 
would  this  give  to  z  (representing  the  amount  of  the  third  kind)  i 

9.  A  merchant  wishes  to  mix  32  pounds  of  tea  at  36  cents 
per  pound,  with  some  at  48  cents,  and  some  at  72  cents. 
How  many  pounds  of  each  kind  must  he  take  to  form  a 
mixture  worth  56  cents  per  pound  ? 

Suggestion.— The  relation  is  2p—x  =  80.  Any  value  for  either 
X  or  y  may  be  taken  which  gives  a  positive  value  for  the  other;  and 
any  positive  value  of  a;  will  give  in  this  case  a  positive  value  for  the 
y,  for2y  =  804-«.  Ifa;  =  4,  y  =  42.  If  a;  =  10,  y  =  45.  If  «  =  1, 
y  =  40^,  etc. 

If  we  add  to  this  example  the  condition  that  the  whole  amount 
of  the  mixture  shall  be  102  pounds,  the  problem  becomes  deter- 
minate ;  as  there  are  then  two  equation?  with  two  unknown  quanti- 
ties. The  equations,  when  reduced,  are  2y—x  =  80,  and  y  +  x  =  70. 
Whence  x  =  20,  and  y  =  50. 

10.  A  man  bought  horses  at  $50  each,  oxen  at  $40,  cows 

at  $25,  calves  at  $10,  so  that  the  average  price  per  head  was 

$30.     How  many  were  bought  of  each  ? 

Suggestions. — The  conditions  tobemetare4r-|-2y=e4-4w,  nega- 
tive and  fractional  values  being  excluded  by  the  nature  of  the  prob- 


334  BUSINESS   RULES. 

lem.  Can  the  conditions  be  met  by  taking  3  cows  and  5  calves  ? 
Why  ?  Can  they  by  taking  2  cows  and  5  calves?  How  ?  Can  they 
by  taking  4  horses  and  6  cows  ?   How  ? 

11.  A  bought  240  barrels  of  molasses  for  ^4,320  ;  worth, 
respectively,  $10,  $14,  $20,  and  $22 ;  how  many  barrels  of 
each  did  he  buy  ?  Ans,,  40,  20,  160,  and  20. 

Suggestion. — The  conditions  are  x+y-\-z  +  w  =  240,  and  4:x-\-2y 
=  z  +  2ic.  Here  are  two  equations  with  four  unknown  quantities, 
hence  any  other  two  conditions  may  be  imposed  which  do  not 
conflict  with  the  nature  of  the  example.  Can  x  =  60  and  y  =  20  ? 
Yes.  This  reduces  the  equations  to  z  +  w  =  160,  andz  +  2w  =  280 ; 
from  which  w  =  120,  and  z  =  40.  Can  the  condition  that  half 
the  quantity  shall  be  of  the  first  two  kinds  be  met  ?  No.  This 
gives  2  +  ^  =  120,  which  subtracted  from  z  +  2w  =  4:X  +  2y,  makes 
w  =  2{2x  +  y)  —  120.  But  since  x  +  y  =  120,  2{2x-\-y)  —  120  >  120. 
.-.  w  >  120,  which  would  make  z  negative. 

12.  I  have  two  kinds  of  molasses  which  cost  me  20  and  30 
cents  per  gallon  ;  I  wish  to  fill  a  hogshead,  that  will  hold 
80  gallons,  with  these  two  kinds.  How  much  of  each  kind 
must  be  taken,  that  I  may  sell  a  gallon  of  the  mixture  at  25 
cents  per  gallon  and  make  10  per  cent,  on  my  purchase? 

Ans.,  58^  of  20  cents,  and  21^^^  of  30  cents. 

13.  A  lumber  merchant  has  several  qualities  of  boards  ; 
and  it  is  required  to  ascertain  how  many,  at  $10  and  $15  per 
thousand  feet,  each,  shall  be  sold  on  an  order  for  60  thousand 
feet,  that  the  price  for  both  qualities  shall  be  $1 2  per  thousand 
feet.    Ans.,  36  thousand  at  $10,  and  24  thousand  at  $15. 

14.  How  many  ounces  of  gold  23  carats  fine,  and  how 
many  20  carats  fine,  must  be  compounded  with  8  ounces  18 
carats  fine,  that  the  alloy  of  the  three  different  qualities  may 
be  22  carats  fine  ?  Ans.,  48  oz.  of  the  1st,  and  8  oz.  of  the  2d. 

[Note. — These  applications  might  be  extended  to  much  greater 
length,  did  space  permit.  The  equation  renders  important  aid  in 
many  problems  in  Compound  Interest,  but  their  discussion  usually 
requires  a  knowledge  of  Quadratics,  and  some  of  them  of  Loga^ 
rithms.     They  must,  therefore,  be  reserved  for  the  future.] 


335 


UA D  R ATI C  ^  :^ aUATIONS 


p 


i^ 


LM>TEI\,  IV< 


fSECJlO¥i  I. 

PURE     QUADRATICS. 

94.  A  Quadratic  Equation  is  an  Equation  of  the  sec- 
ond degree  (6,  8). 

Oo,  Quadratic  Equations  are  distinguished  as  Pure 
(called  also  Incomplete),  and  Affected  (called  also  Complete.) 

90.  A  Pure  Quadratic  Equation  is  an  equation  which 
contains  no  power  of  the  unknown  quantity  but  the  second  ; 
as  ax^-{-b  =  cd,   x^— 3b  =  102. 

97.  An  Affected  Quadratic  Equation  is  an  equation 
which  contains  terms  of  the  second  degree  and  also  of  the 
first,  with  respect  to  the  unknown  quantity  or  quantities; 
as  a:^—4^=  12,  ^xy—x—y"^  =  16«,  mxy  +  y  =  b. 

98.  A  Root  of  an  equation  is  a  quantity  which  sub- 
stituted for  the  unknown  quantity  satisfies  the  equation. 

Note  the  difference  between  this  use  of  the  word  root,  and  its 
former  use  as  defined  in  Art.  39,  Here  we  speak  of  *'  the  root  of 
an  equation,"  meaning  the  value  of  the  unknown  quantity ;  in  the 
former  sense  we  speak  of  *'the  root  of  a  number,"  meaning  one  of 
its  equal  factors.        

99.  Prob.— To  solve  a  Pure  Quadratic  Equation. 
Rule. — /.  Transpose  all  the  terms  containing  the 

unknown  quantity  into  the  first  meniber,  and  unite 
fheTJi  into  one,  clearing  of  fractions  if  vccrsanry, 


336  QUADRATIC    EQUATIONS. 

II.  Transpose  the  known  terms  into  the  second 
inemher. 

III.  Divide  by  the  coefficient  of  the  unknown  quan- 
tity, and  extract  the  square  root  of  each  member. 

Demonstration. — According  to  the  definition  of  a  Pure  Quadratic, 
all  the  terms  containing  the  unknown  quantity  contain  its  square. 
Hence  they  can  be  transposed  and  united  into  one  by  atlding  with 
reference  to  the  square  of  the  unknown  quantity  (16).  Extracting 
the  square  root  of  the  first  member  gives  the  first  power  of  the 
unknown  quantity,  i.  e.,  the  quantity  itself.  And  taking  the  square 
root  of  each  member  does  not  destroy  the  equation,  since  like  roots 
of  equal  quantities  are  equal. 

100,  Cor.  1. — Every  Pure  Quadratic  Equation  has  two 
roots  7imnerically  equal  hut  with  opposite  signs. 

This  is  apparent  since  every  such  equation,  as  the  process  of 
solution  shows,  can  be  reduced  to  the  form  x^  =  a  {a  representing 
any  quantity  whatever).  Whence,  extracting  the  root,  we  have 
X  =  ±  ^a  \  as  the  square  root  of  a  quantity  is  both  +  and  —  (203). 

Scholium. — The  question  naturally  arises,  Why  not  put  the  ambigu- 
ous sign  (the  ±)  before  the  a-,  as  well  as  before  the  second  member? 
It  is  proper  to  ;  but  there  is  no  advantage  gained  by  it.  Thus,  if 
we  write  ±x  =  i^/^,  we  have  -^-x  —  ±'\/a,  or  —  .f  =  ±  y^a.  But 
the  former  is  a?  =  ±  -v/a,  and  the  latter  becomes  so  by  changing  the 
signs  of  both  members.  So  that  all  we  learn  in  either  case,  is  that 
ar  =  +  "V^a,  and  x  =  —  ya. 

101.  Cor.  2. — The  roots  of  a  Pure  Quadratic  Equation 
may  loth  he  imaginary,  and  both  ivill  he  if  one  is.  For  if 
after  having  transposed  and  reduced  to  the  form  x^=za,  the 
second  member  is  negative,  as  x^=^  —«,  extracting  the  square 
root  gives  x=-\-^/ —a,  and  x=-~^/ —a,  both  imaginary. 


EXAMPLES. 

Ex.  1.  Given  3r>«- 

-10— .^2=12  + 4^:2-54  to  find  the  value  of  a:. 

Model  Solution. 

Operation. 

zx'-n-x'  ^^n+^x'-u, 

(2) 

_2.r^  =  -32, 

(3) 

«''  =  16,        0?  =  ±4, 

PURE   QUADRATICS.  337 

Explanation. — Transposing  and  uniting  terms,  I  liave  —2x*.—  —32. 
Dividing  each  member  by  —2,  I  have  x'  =  16.  Extracting  the 
square  root  of  each  member,  I  find  a;  =  ±4  (read,  ''x  equals  plus 
and  minus  4  "). 

Veriflcation. — Substituting  +4  for  x,  the  equation  becomes  48 
—  10-16  =  12  +  64—54,  or  22  =  22.  Substituting  —4  gives  just 
the  same  since  —4  squared,  (—4)'^,  is  the  same  as  +4  squared. 

2.  Given  a^-\-l  =  ~-\-4:,to find x.        Roots,  x=  ±2. 

4 

3.  Given  -^^-^ — -  =  — ^ — ,  to  find  x. 

10  o 

Roots,  x=  ±3. 

5  5  8 

4.  Given 1-  -; =  - ,  to  find  the  values  of  x. 

4  +  a:       4:—x       3 

Roots,  X  =  ±1, 

6.  Given  a^—ah  =  d,  to  find  the  values  of  x. 

Roots,  x=  ^^d-\-db. 

6.  Given  |^_3  +  ^  =  ^-;c2+^,  to  find  the  values 
of  a;.  Roots,  x—  ±3, 


7.  Given  13— v3ar5+16  =  5,  to  find  the  values  of  x. 

Roots,  a;  =  ±4. 

8.  Given  x  -f-  Vx^ + «  =  —, ,  to  find  the  values  of  x. 


Roots,  X  =  iiVSa. 


15 


9.  Given  --=^  =x-\-w^-\-^.         Roots,  x  z=  ±2, 
10.  Given  V^+ri^-^i^^.  Roots.  x^^/2^{^h)^^. 


338  QUADBATIC  EQUATIONS. 

11.  GiTen =  ax.  Roots,  x=  +  — — 

12.  Given  ^-^=?^^-». 

X  XX 

Boots,  x=  ±(Vm—Vn), 

13.  Giyen^,--+1  =  --1  +  -.  

Roots,  X  =  ±\/a^—^, 

14.  Given  12— rc^  i^x^::  100 :  25.        Boots,  x=z  ±2. 


Operation.     12— x^ :  ^a;' : 


100 :  25 


12-x^ : «"  : :  50  :  25  : :  2 : 1 
13:rB»::3:l 
4:35"  ::  1:  1     .:  x  =  ±2. 

[Note.— Use  the  principles  of  proportion  in  solving  these.] 
16.  Given  j^x^-^-^x^S  :  ^x^-.:^x^^s  : :  9  :  3. 

Boots,  X  =  :t  4'v/2. 

16.  Given  i{x^—5f :  x^—6  : :  2 : 1.       Boots,  x  =  ±S. 

Operation.  i(x^-5y  :  a;'-5  : :  2  : 1 
«'— 5  : 1  : :  4  : 1 
«'~5:5  ::4:5 

«» :  5  : :  9  : 5    .-.  aj  =  ±3. 

17.  Given  |(11  +a;2) :  \{4x^—2)  : :  5  : 2. 

Boots,  a;  =  ±2. 

18.  Given  a;2+ 4 :  a;2— 11  : :  100 :  40. 

Boots,  X  =  ±  \/2T. 

19.  Given  ^±^  =  2  -  ^+^.    Roots,  x  =  ±2. 

20.  Given  ^—^  +  -^-^^  =  _.  Boots,  x  =  ±8. 

21.  Given  (a;4-2)2  =  4:X-{-6.  Boots,  x  =  ±1, 

22.  Given  i{x'i-12)  =  ix^-1.  Boots,  x  =  ±6, 

23.  Given  ^  +  ^         ^"^  ^ 


x^—7x      x^  +  Hx      x^—7d 

Boots,  X  =  ±9, 


PURE    QUADRATICS.  339 


24.  Given  a;  + V^a/2— 4a;  =  1.       Roots,  x  =  ±f 


25.  Given  Va  +  x  =  Vx+Vo^-^-^.  

Roots,  x=  ±Va^—i^. 


26.  Given  Vl  +  x^-h^l+a^+Vl  —  3^=  Vl—x^. 
Verify.  Roots,  x  =  ±|\/^^. 

27.  Given  ^¥T^  =  Vo^T^. 

Roots,  x=  ±^VHl^-a% 


28.  Given*— VaHa^^ 


Roots,  X  =  ±-^^3-^^. 


29.  Given  \ =  *.  Roots,  x  z=z  ±- — -. 


30.  Given  X^^±^±l  =  3.  Roots,  x  =  ±2. 


31.  Given  vWl     Vg^  ^  ^   ^^^^     ^  ^  ±iV5. 

32.  Given 7=^  H f=^  =  x- 

x^\/2—xi      x-V2-x^ 

Roots,  X  =  ±  Vs. 
1  la 


33.  Given 


a—  Va^—3^      a  +  Va^  -a:^       ^ 

Roots,  X  =  ±iaV3. 


«>.     r^-        Va^+x^-\-x       b         „    ^  ,  a(b—c) 

34.  Given  \ __----  =  -•       Roots,  x  =  ±  -— 7^- 


340  QUADRATIC    EQUATIONS. 

35.  Giyen  — =L +  —== —  =  ^  • 


Boots,  X  =  ±  jVs. 


36.  Given =  a  — 


l  +  x-i-Vl-i-x^  l—x-\-Vl+x^ 


Roots,  x=  ±V{a—2y—l, 


37.  Given ; H ; =  ax. 

x-\-V2—a^      x—V2-^x^ 


Roots 


■      A  +  1 


38.  Given  - —  +  -!—  =  a. 

Vl— a;+l       vl  +  ^— 1 


Roots. 


.-±V'4- 


APPLICATIONS. 

Ex.  1.  What  two  numbers  are  those  whose  sum  is  to  the 
greater  as  10 :  7,  and  whose  sum  multiplied  by  the  less  pro- 
duces 270?  Ans.,  21  and  9. 

Suggestion. — Let  10a?  =  the  sum  of  the  numbers,  and  7a?  the 
greater. 

Scholium. — It  is  customary  to  omit  the  negative  roots  in  giving 
answers  to  examples,  the  nature  of  which  renders  such  answers 
impossible.  In  this  case  the  question  is  about  pure  number,  and 
hence  the  answers  should  be  given  without  signs. 

2.  There  are  two  numbers  whose  ratio  is  that  of  4  to  5, 
and  the  difference  of  whose  squares  is  81.  What  are  the 
numbers?  Ans.,  12  and  15. 

3.  What  two  numbers  are  those  whose  difference  is  to  the 
greater  as  2  to  9,  and  the  difference  of  whose  squares  is  128  ? 
Verify. 


PURE    QUADRATICS.  341 

4.  Find  three  numbers  which  bear  the  same  ratio  to  each 
other  as  J,  |,  and  J  do  to  each  other,  and  the  sum  of  whose 
squares  is  724.  Numbers,  12,  16,  18. 

6.  Find  three  numbers  in  the  ratio  of  m,  u,  and  p,  the 
sum  of  whose  squares  is  equal  to  a. 

Numbers,  ±\/    »  ,  ^  ,    »,  ±\/    .  .  ^t  .    .,  and 


V  rn^  +  n 


pi 


6.  Divide  14  into  two  parts  so  that  the  greater  part  divid- 
ed by  the  less  shall  be  to  the  less  divided  by  the  greater  as 
16  to  9. 

Suggestion. — Having  — —   :  : :  16  :  9,  it  follows  that 

14 — X         X 

x" :  {U~xy  : :  16  :  9,  and  x :  14-x  : :  4  :  3,   and  a; :  14  : :  4  :  7.     .'.  x 

=  8,  and  14— x  =  6. 

7.  Divide  a  into  two  parts  so  that  the  greater  part  divided 
by  the  less  shall  be  to  the  less  divided  by  the  greater  as 
w  to  n. 

Parts,  -—= ;=,  and 


Scholium. — Example  7  is  example  6  generalized.  Tl^  pupil 
should  deduce  the  results  in  the  former  from  these.     Thus,  sub- 

stftuting  a  =  14,  m  =  16,  and  n  =  9,       _/ — —^=  8,  etc. 

8.  What  two  numbers  are  they,  whose  product  is  126,  and 
the  quotient  of  the  greater  divided  by  the  less,  S^  ?  Gener- 
alize this.  Ans.,  6  and  21. 

9.  The  sum  of  the  squares  of  two  numbers  is  370,  and 
the  difference  of  their  squares  208.     Required  the  numbers. 

Numbers,  9  and  17. 


342  QUADEATIC   EQUATIONS. 


10.  Generalize  the  9th,  and  show  that  Jv2(5  f^)  and 
^V'^(s—d)  are  general  results. 

11.  For  comparatively  small  distances  above  the  earth's 
surface  the  distances  through  which  bodies  fall  under  the 
influence  of  gravity  are  as  the  squares  of  the  times.  Thus, 
if  one  body  is  falling  2  seconds  and  another  3,  the  distances 
fallen  through  are  as  4:9.  A  body  falls  4  times  as  far  in  2 
seconds  as  in  1,  and  9  times  as  far  in  3  seconds.  These  facts 
are  learned  both  by  observation  and  theoretically.  It  is 
also  observed  that  a  body  falls  IGy^^^  feet  in  one  second. 
How  long  is  a  body  in  f alhng  500  feet  ?  One  mile  (5280  ft.)  ? 
Five  miles  ?  Ans.,  To  fall  500  ft.  requires  5.58  seconds. 
To  fall  5  miles  requires  40.51  seconds. 

12.  A  and  B  lay  out  some  money  in  a  speculation.  A 
disposes  of  his  bargain  for  111,  and  gains  as  much  per  cent, 
as  B  lays  out.  B  succeeds  in  gaining  $36 ;  and  it  appears 
that  A  gains  four  times  as  much  per  cent,  as  B.  Eequired  the 
capital  of  each.  Results^  $5  =  A's  capital,  and  $120  =  B's. 

13.  A  money  safe  contains  a  certain  number  of  drawers. 
In  each  drawer  there  are  as  many  divisions  as  there  are 
drawers,  and  in  each  division  there  are  four  times  as  many 
dollars  as  there  are  drawers.  The  whole  sum  in  the  safe  is 
$5,324;  what  is  the  number  of  drawers?  Afis.,  11. 

14.  Two  travelers,  A  and  B,  set  out  to  meet  each  other; 
A  leaving  the  town  C  at  the  same  time  that  B  left  D.  They 
travelled  the  direct  road  from  C  to  D,  and  on  meeting  it 
appeared  that  A  had  travelled  18  miles  more  than  B;  and 
that  A  could  have  gone  B's  journey  in  15|  days,  but  B  would 
have  been  28  days  in  performing  A's  journey.  What  is  the 
distance  between  C  and  D  ?  Ans.,  126  miles. 


C  M  D 

'- 1         .  '•  ^ 1 


If  x  =  CM  =  the  distance  A  travelled,  then  «— 18  =  MD  =  the  dis- 


PURE    QUADRATICS.  343 

tance  B  travelled.  =  distance  A  travelled  a  day ;   and  — 

15^  28 

=  distance  B  travelled  a  day.     Notice  that  the  times  are  equal. 

15.  From  two  places  at  an  unknown  distance,  two  bodies, 
A  and  B,  move  toward  each  other  till  they  meet,  A  going 
a  miles  more  than  B.  A  would  have  described  B's  dis- 
tance in  n  hours,  and  B  would  have  described  A's  distance 
in  m  hours.      What  was  the  distance  of  the  two  places 

from  each  other  ?  ^/m-\-  ^/n 

Ans.,  a  X  -pr yr- 

16.  A  and  B  engaged  to  work  for  a  certain  number  of 
days.  A  lost  4  days  of  the  time  and  received  $18.75. 
B  lost  7  days  and  received  $12.  Now  had  A  lost  7  and 
B  4  days,  the  amounts  received  would  have  been  equal. 
How  long  did  they  engage  to  work  and  at  what  rates  ? 

Ans.,  Whole  time,  19  days. 

Suggestion. — ^Ifa;=  the  whole  time,  what  represents  A's  daily 
wages  ?  What  B's  ?  After  the  equation  is  formed,  see  if  you  can- 
not strike  out  a  numerical  factor  from  both  members,  and  extract 
the  root  without  expanding. 

17.  A  vintner  drew  a  certain  quantity  of  wine  out  of  a 
full  vessel  that  held  256  gallons  ;  and  then  filled  the  vessel 
with  water,  and  drew  off  the  same  number  of  gallons 
as  before,  and  so  on  for  four  draughts,  when  there  were 
only  81  gallons  of  pure  wine  left.  How  much  wine  did 
he  draw  each  time  ?        Am.,  64,  48,  36,  and  27  gallons. 

Suggestion. — If  he  drew  out  -  part  of  the  contents  of  the  cask 

x—l 
each  time,  there  remained  after  the  first  drawing th    of   the 

x—l      X 1       (x—iy 

wine ;  after  the  second  x  or —,  and  after  the  fourth 

X  X  Q?       ' 

{x-\y       (x-i)*    ,,       x-\     «     „„        11 

S^-    ••-V-=^3^^'«'-^=f     Whence -  =  ^. 


S41  QUADRATIC    EQUATIONS. 

18.  A  number  a  is  diminished  by  the  nth.  part  of  itself, 
this  remainder  is  diminished  by  the  nth  part  of  itself, 
and  so  on  to  the  fourth  remainder,  which  is  equal  to  h. 
Required  the  value  of  n.  ^^ 

A71S.,    - 


19.  There  is  a  number  such  that,  if  the  square  root  of 
three  times  its  square  +  4,  be  taken,  the  quotient  of  this 
root  increased  by  2,  divided  by  the  root  diminished  by  2,  is  3. 
What  is  the  number  ? 

Query. — Which  of  the  equations  in  the  preceding  j^art  of  this 
section  does  this  give  rise  to  ? 

20.  If  the  square  root  of  the  difference  between  the 
square  of  a  certain  number  and  2,  be  both  added  to  and 
subtracted  from  the  number  itself,  the  sum  of  the  recip- 
rocals of  the  result  is  ^  of  the  number  itself.  What  is  the 
number  ? 

Query. — Which  of  the  equations  in  the  preceding  part  of  this 
section  does  this  give  rise  to  ?     With  what  modifications  ? 


AFFE   f  CTED, 
J^^TM>N  II. 


102.  An  AfiFected  Quadratic  equation  is  an 
equation  which  contains  terms  of  the  second  degree  and 
also  of  the  first  with  respect  to  the  unknown  quantity. 


AFFECTED    QUADRATICS.  345 

x^-dx  =  12,  ^  +  Sax^  =  ??£+^^    ^^^^  cM  __  ^^^  ^  g^ 

5  0^ 

—  0,  are  affected  quadratic  equations. 

103.  Prob.— To  solve  an  Affected  Quadratic  Equation. 
Rule. — /.  Reduce  the  equation  to  the  form  x^+ax 

//.  Add  the  square  of  half  the  coefficients  of  the 
second  term  to  each  mernher  of  the  equation. 

III.  Extract  the  square  root  of  each  member,  thus 
producing  a  simple  equation  from,  which  the  value  of 
the  unhnown  quantity  is  found  by  simple  transposition. 

Demonstration. — By  definition  an  Affected  Quadratic  Equation 
contains  l>ut  tbree  kinds  of  terms,  viz.:  terms  containing  the  square 
of  the  unknown  quantity,  terms  containing  the  first  power  of  the 
unknown  quantity,  and  known  terms.  Each  of  the  three  kinds 
of  terms  may,  by  clearing  of  fractions,  transposition,  and  uniting,  as 
the  particular  example  may  require,  be  united  into  one,  and  the 
results  arranged  in  the  order  given.  If,  then,  the  first  term,  i.  e.  the 
one  containing  the  square  of  the  unknown  quantity,  has  a  coeffi- 
cient other  than  unity,  or  is  negative,  its  coefficient  can  be  rendered 
unity  and  positive  without  destroying  the  equation  by  dividing 
both  the  members  by  whatever  coefficient  this  term  may  have 
after  the  first  reductions.     The  equation  will  then  take  the  form 

T}±ax=  ±h.    Now  adding  (-)   to  the  first  member  makes  it  a 

perfect  square  (the  squareof  a;± -),  since  a  trinomial  is  a  perfect 

square  when  one  of  its  terms  (the  middle  one,  a«,  in  this  case)  is 
±  twice  the  product  of  the  square  roots  of  the  other  two,  these  two 
being  both  positive  (123,  Part  I).  But  if  we  add  the  square  of 
half  the  coefficient  of  tlie  second  term  to  the  first  member  to  make 

*  The  characteriBticB  of  this  form  are,  that  the  first  member  consists  of  two 
terms,  the  first  of  which  is  positive  and  contains  simply  the  square  of  the  unknown 
quantity,  its  coefficient  being  unity,  while  the  second  has  the  first  power  of  the 
unknown  quantity,  with  any  coefficient  (a)  positive  or  negative,  integral  or  frac- 
tional ;  and  the  second  member  consists  of  known  terms  (b). 


346  QUADRATIC    EQUATION'S. 

it  a  complete  square,  we  must  add  it  to  the  second  member  to  pre- 
serve the  equality  of  the  members.  Having  extracted  the  square 
root  of  each  member,  these  roots  are  equal,  since  like  roots  of  equals 
are  equal.     Now,  since  the  first  term  of  the  trinomial  square  is  x\ 

and  the  last  i  —  J  does  not  contain  x,  its  square  root  is  a  binomial 

consisting  of  x  ±  the  square  root  of  its  third  term,  or  half  the  co- 
eflBcient  of  the  middle  term,  and  hence  a  known  quantity.  The 
square  root  of  the  second  member  can  be  taken  exactly,  approxi- 
mately, or  indicated,  as  the  case  may  be.  Finally,  as  the  first  term 
of  this  resulting  equation  is  simply  the  unknown  quantity,  its  value 
is  found  by  transposing  the  second  term. 


EXAMPLES. 

X^  x^ 

Ex.  1.  Given  6^x—^  =  x—2l-\--,  to  find  the  value  of 

o  o 

X,  and  verify. 

Model  Solution— Operation. 

48_8a;-a!'  =  8aj-184-«' 
-2x''-16x  =  -66 

x''  +  Sx  =  dS 
a;''  +  8a;+16  =  33  +  16  =  49 

X  +  4:  =  ±7, 

a?  =  ±  7—4  =  3,  and —11 

Explanation.— Clearing  the  equation  of  fractions,  transposing  and 
uniting,  and  dividing  each  member  by  —2,  I  have  x^  +  Sx  =  33. 
Now  since  Sx  contains  the  square  root  of  a?"  as  one  of  its  factors,  the 
other  factor,  8,  is  twice  the  square  root  of  the  other  term  of  a  trino- 
mial square  (123,  Part  I).  Hence  ^  of  8  squared  (16)  is  the  third 
term.  Adding  this  term  I  have  x^  +  Sx  + 16,  which  is  a  perfect  square. 
But  as  I  have  added  16  to  the  first  member  to  make  it  a  perfect 
square,  I  must  add  it  to  the  second  member  to  preserve  the  equality. 
This  gives  a;-  +  8a;  +  16  =  49.  Extracting  the  square  root  of  both 
members,  I  have  a:  +  4  =  ±7.  Finally,  transposing  the  4, 1  have 
x  =  —4  ±7,  i.  6.  a?  =  3  and  —11.  Both  are  correct.  Hence  there 
are  two  roots  (values  ofx)  of  this  equation,  3  and  —11. 


AFFECTED    QUADRATICS.  347 

Verlflcatlon. — To  verify  the  value  a;  =  3, 1  substitute  3  for  a;  in  the 
given  equation,  and  have  6-3— |  =  3— 2J+|,  or  3— |  =  |4-|,  or 
ys.  ^  y.  To  verify  the  value  j^  — ll,l8ubBtitutefor«,  — 11,  in  the 
given  equation,  and  have  6  +  11— J^f^  =  -ll-2|  +  if-i-,  or  17— »-fL 
=  -131  +  ifi,orV=¥- 

2.  Given  x^—Sx-\-5  =  14,  to  find  the  values  of  x,  and 
verify.  Result,  re  =  9,  and  —1. 

3.  Find  the  roots  of  the  equation  a:2_i2a:-f  30  =  3,  and 
verify.  Boots,  9  and  3. 

4^; 9 

4.  Find  the  roots  of  a:— 2  = • 

X 

Suggestion.— This  reduces  to  x'—Qx  =  -  9.  Whence,  completing 
the  square,  a:'-6a;  +  9  =  9-9  =  0,  and  a:-3  =  0,  or  x  =  3.  In  this 
case  it  appears  that  the  equation  has  but  one  root. 

6.  Giyen  i  =  ^^^—-^ ,  to  find  the  values  of  x. 
5  X 

6.  Find  the  roots  of  3(a:— 4)  =  — i^^J. 

^  X 

Suggestion.-  This  reduces  to  x'—^x  = — 20.  Whence,  completing 
the  square,  a;'— 8./ +  16  =  16—20  =  —4,  and  extracting  the  root, 
x— 4  =  ±  2^/— li  or  «  =  4  ±  2\/~^,  two  imaginary  roots. 

7.  Find  the  values  of  a;  in  -  =  — -z* 

5       X — 0 


Results,  X  =  5  +  2\/— 5,  and  5— 2a/^. 


8.  Find  the  roots  of  x^^^  +  n^  =  3— J5?. 

X  X 


Roots,  —30,  and  —40. 


9.  Find  the  roots  of  i(?|±l)  =  ?_a:. 
"Zx  X 


Only  one  root,  —2. 


348  QUADRATIC    EQUATIOXS. 

10.  Find  the  roots  of  7(a:+7)  +  ^^^?^— ^  =  0. 

Roots,  5(— 1  +  a/— 1),  and  5(— 1  — V— 1),  or  as  the 
same  may  be  written,  —5(1  — a/— 1),  and  —  5(1  -f  a/— 1). 

11.  Find  the  roots  of  ^x^—^x  +  %^  =  42|,  and  verify. 

Boots,  7,  and  —6  J. 

Scholium. — This  process  of  adding  the  square  of  half  the  co- 
efficient of  the  first  power  of  the  unkno^vn  quantit}^  to  the  first 
member,  in  order  to  make  it  a  perfect  square,  is  called  Completlng 
THE  Square.  There  are  a  variety  of  other  ways  of  completing  the 
square  of  an  afifected  quadratic,  some  of  which  will  be  given  as  we 
l^roceed ;  but  this  is  the  most  important.  This  method  will  solve 
all  cases :  others  are  mere  matters  of  convenience,  in  special  cases. 

12.  Given  x^—x-\-3  =  45,  to  find  its  roots,  and  verify. 

13.  Given  2ic2-|-8a;— 20  =  70,  to  find  its  roots,  and 
verify. 

14.  Given  5 —  =  Id^x — -  ,  to  find  its  roots, 

and  verify. 

25 ^/^ 

15.  Given  6x-\ =  44,  to  find  the  values  of  x. 

X 

Suggestion. — Clearing  of  fractions,  Qx^  +  ^S—'Sx  =  44a;, 
Transposing  and  uniting,  Qx'^—4:7x  =  —35, 
Dividing  by  6,  x^—^x  =  — ^, 

Completing  the  square,  x^— Ya;  +  (||)'  =  {iW—^  =  tW^» 
Extracting  the  root,  x—^  =  ±  ||, 
Transposing,  a?  =  ff  ±  f|^  =  7,  and  |. 

16.  Find  the  roots  of  bx =  2x-\ — 

Suggestion. — Notice  the  compound  negative  term.  Cleared  of 
fractions  and  reduced,  the  equation  becomes  x'^—Sx  =  4.  .*.  «  =  4, 
and  ~1. 

17.  Find  the  roots  of ^-^ —  =  3. 

X  4:X'^ 

Suggestion. — Multiply  by  4x^  to  clear  effractions. 


AFFE(TKD    QUADRATICS.  349 

18.  Find  the  roots  of  Sxi—20x—6'Z  =  7:i'— 2.^2  + 100. 

Boots,  9,  and  —  3|. 

19.  Find  the  roots  of  2a;— 2  =  2  +  -. 

X 

Roots,  3,  and  —1. 

20.  Find  the  roots  of  ix^^^x  +  l  =  8-|. 

Roots,  I,  and  —  |. 

21.  Find  the  values  of  x  in    the    equation    — —^  +5 
^^^  Result,  X  =  5|,  and  5. 


a;  +  10 


22.  Find  the  values  of  x  in  the  equation  — -^  —  —  =  6. 

Result,  X  =  10,  and  —  f. 

23.  Giveni(a;-f4)-^  =4(4a;H-7)-l,    to   find   the 

X-~~o 

values  of  x.  Restdt,  x  =  21,  and  5. 

24.  Given  '"^^  +  — f^  =  5^,  to  find  the  roots,  and 

X  X-f-  L4 

verify. 

25.  Find  the  roots  of  ^(8— a;)— ^""      =  i(^  —  2),  and 

X'^o 

verify. 

,      ,2a?+9       4a;— 3       ^  .  3a;— 16        , 

26.  Find  the  roots  of  — ^  +  j— 3  =  3  +  --^—,  and 

verify. 

27.  Find  the  values  of  x  in  the  equation  3a;2_2aa;  =  b. 


J,      ;,            a±Va'-^Sb 
Result,  X  = -^ • 


350  QUADRATIC    EQUATIONS. 

Operation.  Sx^—2ax  =  h; 

Dividing  by  3,  x'-^x  =  l, 

o  3 

Completing  the  square,  «— ^«  +  (q)   =^  +  -  =  , 

3        \3/        9        3  9 


Extracting  the  square  root,  x—-  =  ±  -^a^  +  3&, 

o  3 

m             .                            a     a/o^  +  36       a±  a/«'  +  3J 
Transposing,  x  =  -  ±  -^- — -- — ,  or ^ — ——. 


Scholium.— The  form  — ^^ signifies  that  there  are  two 

o 

values  of  «;  i.  e.,  that  each  of  the  signs  +  and  —  may  be  used.  Thus 
the  values  in  this  case  are 

X  = ^ ,  and  X  = ^ 

3  '  3 

28.  Find  the  roots  of  Sx^-\-6ax  =  m. 

Roots,  == — 

o 

29.  Find  the  roots  of  ^a^¥x^—Qa%^x  =  ¥. 


Roots, 


ZaW 


X        (I        2 

30.  Find  the  yalues  of  x  in  the  equation  -  +  -  =  -. 

a       X      a 

Result,  X  =  1±a/1— ^. 

31.  Find  the  roots  of  ^:Z^  =  t 

Sa—2x        4 

Roots,  ia,  and  ^a. 

104,  Cor.  1. — An  affected  quadratic  equation  has  tivo 
roots.  These  roots  may  hoth  he  positive,  loth  he  negative,  or 
one  positive  and  the  other  negative.  They  are  hoth  real,  or 
hoth  imaginary. 

Demonstration. — Let  x'^-\-px  —  qhe  any  affected  quadratic  equa- 
tion reduced  to  the  form  for  completing  the  square.  lu  this  form 
p  and  q  may  be  either  positive  or  negative,  integral  or  fractional, 


AFFECTED    QUADRATICS.  351 

Solving  this  equation  we  have  a?  =  — |  ±  a/^  -\-q.     We  will  now 

observe  what  diflFerent  fonns  this  expression  can  take,  depending 
up(jn  the  signs  and  relative  values  of  y>  and  q. 

Ist.    Wh^i  p  and  q  are  both  positive.     The  tfigns  will  then  stand  as 

given  ;  i.  e.,  x  =  —^±  a/ ^^ +q.     Now,  it  is  evident  that  i/  i"  +  2 

/    a 

>^,  for  4/^+2  is  the  square  root  of  something  more  than 
^.     Hence,  as  ^  <  y—  +J,  — |  +  y-  +  5'  is  positive;  but  -^ 

—  i/ J-  +q  ia  negative,  for  both  parts  are  negative.     Moreover  the 

negative  root  is  numerically  greater  than  the  positive,  since  the 
former  is  the  numerical  sum  of  the  two  parts,  and  the  latter  the 
numerical  difference.  .*.  When  p  and  q  are  both  +  in  the  given 
Ibrm,  one  root  is  positive  and  the  other  negative,  and  the  negative 
root  is  numerically  greater  than  the  positive  one.  See  Example  1, 
above. 

3nd.     When  p  is  negative  and  q  positive.     We  then  have  x= ^ 


radical,  x  is  positive ;  but  if  we  take  the  —  sign,  a;  is  negative,  since 


/? 


+  q>~.      Moreover,   the  positive  root  is  numerically   the 

greater.  .  *.  When  p  is  negative  and  q  positive,  one  root  is  positi  ve  and 
the  other  negative ;  but  the  positive  root  is  numerically  greater  than 
the  negative.     See  Example  2,  above. 

8rd.   When  p  avd  q  are  hoth  negative.    We  then  have  a?  = ^ 


-  \/^=f^  (-«)  =  f  -  /:--?•  ^  «>»  iff'  >  Vr  -  * '« 

real,  and  as  it  is  less  than  ^,  both  values  are  positive.    See  Ex.  8.   If 

^  =  J,  4/^ — q  —  0,  and  there  is  but  one  value  of  «.  and  this  is  posi- 

tive.     (It  is  customary  to  caU  this  two  equal  positive  roots,  for  the 
sake  of  analogy,  and  for  other  reasons  which  cannot  now  be  appre- 


352  QUADRATIC    EQUATIO]SrS. 


dated  by  the  pupil.)    See  Examples  4  and  5.    If  ?•  <  ?,  i/ j  ~  2 

becomes  the  square  root  of  a  negative  quantity  and  hence  imaginary. 
See  Examples  6  and  7. 

4th.    When  p  is  positive  aiid  q  negative.     We   then  have  x  =  — ~ 

±  |/  ^ — q.  As  before,  this  gives  two  real  roots  when  ?  <  j-  •  When 
this  is  the  case  both  roots  are  negative.  [Let  the  pupil  show  how 
this  is  seen.]     When  2  =  V  ,  the  roots  are  equal  and  negative ;  **.  e. 

there  is  but  one.  When  y  <  q  both  roots  are  imaginary.  See 
Examples  8,  9,  and  10. 

[Note. — It  is  not  important  that  the  pupil  remember  all  these 
forms;  but  it  is  an  excellent  exercise  to  give  the  discussion.  The 
ingenious  student  can  put  the  results  in  a  very  neat  analytical  table.] 

Scholium. — It  may  be  asked  why,  when  we  extract  the  square 

p^  p^ 

root  of  each  member  of  the  equation  x'^+px  +  j-  =  q+^  ,  we   write 

the  ambiguous  sign  only  before  the  root  of  the  second  member.  The 
reason  is  the  same  as  given  under  the  Pure  Quadratics,  Art.  100, 
Scholium.  Thus,  in  strict  propriety,  the  square  root  of  each  mem- 
ber  of  this  equation   being  taken  or    indicated    gives  ±  l^+f ) 

=  "^  \/  j-^q-    But  take  these  signs  in  any  order  we  can,  it  amounts 

to  taking  the  roots  as  having  like  signs  (both  + ,  or  both  — )  or 
unlike  signs  (one  +  and  the  other  — ).  Hence  it  is  sufficient  to  give 
the  ambiguous  sign  to  one  member  only ;  and  it  is  most  convenient 
to  give  it  to  the  second. 

32.  Given  V^  +  5  x  ^/x^\^l'^=l'^,  to  find  the  values  of  x. 

Result y  a:  =  4,  and  —21. 
Suggestion. — First  clear  the  equation  of  radicals. 


33.  Given  A/4a;+5x\/7a;  +  l  =  30,  to  find  the  values 
of  X.  Vahies,  x  =  6,  and  — 6^. 

34.  Given  x-\-6  =  \/x-\-6-\-6,  to  find  the  values  of  a: 

VahteSf  x  =  4,  and  —1. 


AFFECTED    QUADRATICS.  363 


35.  Given  x-\-16—7Vx-\-W=10—Wx-\-16,   to  find 
the  values  of  x.  Vaiucs,  ar  =  9,  and  —12. 


Suggestion.— Put   the  equation  in  the  form  x+6  =  ^^x+lQ, 
and  then  square. 


36.  Given  3Vx+6-^2  =  x-^Vx-^(),  to  find  the  vahic-- 
of  X.  Result,  X  =  10,  and  —2. 

1         4 

37.  Given  y-^-  =  -— :,  to  find  the  roots. 

ResuUy  x=  VS,  iV3. 

4 
Suggestion. — Multiply  by  y  and  transpose,  and  y' :ry  =  —  1. 

4        4      4  1 

Completing  the  square,  y' ~y+~  =  -_l  =         Extracting  the 

and  /\/|  =  <y/3,  and  ^\^^. 

10f>,  Cor.  2. — An  affected  quadratic  being  reduced  to  the 
form  x2  -f  px  =  4,  i\\e.  value  of  x  is  half  the  coefficient  of 
the  second  term  taken  ivith  the  opposite  sign,  ±  the  square  root 
of  the  sum  of  the  square  of  this  half  coefficient,  and  the  knoum 
term  of  the  equation.     This  is  observed  directly  from  the  form 

X  =  — ^  iA  /  j-  +  q,  dnd  more  in  detail  in  the  demonstration 

of  the  preceding  corollaiy. 

[Note. — The  pupil  should  use  this  method  in  practice,  but  be 
careful  that  the  complete  method  and  its  full  demonstration  is  not 
lost  sight  of.] 

38.  Write  out  the  roots  of  the  following  without  going 
through  the  operations  of  completing  the  square,  etc.; 
z^-{-4:Z  =  eO;  y2_4y  — 60;  x^-f  16a- =  — 60;  .^2  —  16a: 
=  -60. 

39.  Reduce  the  following  to  the  form  x^-{-px  =  q,  and 
then  write  out  the  roots  as  above : 


354  QUADEATIC    EQUATIONS. 

Sx^-i-2x-\-6  =  11,  gives  x^  +  ^x  =  J.     Whence  a;  =  — J 
±  V4+1  =  -i±i  =  h  and  -|. 
—-  +  -—=  —4,  gives  x^-\-~x  =z  — — .    Whence  x  = 

10  /lOO       64       -10±6  ,,        ^       ^, 

-■3"  ±  V -9-  -  y  =  —3—  ="  -^*'  ^^^  -"*• 

4a; =  14,  givesa;2— -2;  =  7.      Whence    x  =  - 


y^g  +  7=:i-^  =  4,and-lf. 


9  ±23 
8 


40.  Given  9x-^—12x-^  =  —3,  to  find  the  values  of  x. 

?,  X  =  3,  and  1. 


Suggestion.    9ar-''  =  — 

X 

1       .        1 

y  +  -      1  +  - 

41.  Given  -|  + 1  =  ^,  to  find  the  roots. 

^     y  y 

Suggestion. — Reduce  the  complex  fractions  to  simple  ones  by 

multiplying  numerator  and  denominator  by  y.   Whence  ^ — -  -{-- — 

y  -1  2/-1 
=  '■^.  Multiply  by  4(y'-l),  and  4yV4  +  4y^  +  8y  +  4  =  ISy'-ld, 
.'.  y  =  3,  and  -\. 

12a 


42.  Given  \/a-\-x-j-V(t—x  =  — ,  to  find  x. 

5'\/a-j-x 

Result,  x  =  —,  and  —  • 
o  0 

43.  GivenV^4-«— V^+^  =  '\/2x,  to  find  the  roots. 

Result,  x=  -  ^±-  _}_  jV2a2  4-2Z^. 

Suggestion,     \/x  +  a—\/2x=:^/x+b.    x+a—2^2x^^2ax-\-2x 

-  x+i:    -2^JW^'2a^  =  J)—a-2x.    ^x^  +  ^a^  =  y^-^^db—^bx  +  a' 
+  ^ax  +  4:X^. 


AFFECTED    QUADRATICS.  356 

44.  Given  -: -—(a^—b^)x  = -. r,  to  find 

a^  +  b^  (a^)-*+(fl2Z>)-i 

the  values  of  x.  Result,  x  =  a,  and  —b. 

Suggestion.     -. t=—, r.   Whence  the  given  equa- 

tion  becomes  x''  —  (a—b)z  =  ah. 

45.  Given  — -= H — — —=  =  — -= ,  to  find 

VX-\-ya — X       \x — ya — x       \x 

the  roots.  Roots,  x  =  -^^ — 


Suggestion. — Add  the  fractions  in  the  first  member,  and 
=  — -.    Whence  2a;'  =?  2fea;-aft. 


2x—a 


106,  Cor.  3. —  Upon  the  principle  that  the  middle  term  of 
a  trinomial  square  is  turice  the  product  of  the  square  roots  of  the 
other  tux)  (94,  95,  Part  I. ),  u)e  can  often  complete  the  square 
mxyre  advantageously  than  by  the  regular  rule. 

46.  Solve  ^7?-\-Ux  =  m. 

Solution. — Dividing  16a;  by  twice  the  square  root  of  4a;',  i.  c,  by 
4a!,  and  adding  the  square  of  the  quotient,  (4)%  to  each  member, 
4a;'  +  16x4-16  =  49.  Extracting  the  root,  2x+4  =  ±  7.  ..  x  =  f, 
and  -V. 

47.  Solve  8a«-12a;  =  36. 

Suggestion- — Divide  the  equation  by  2,  and  proceed  as  above. 
4^-  — 6^  +  (f)'  =  V-.  2.r— f  =  ±1,  and  x  =  3,  and  -If  In  thia 
example  the  regular  method  is  better. 

48.  Solve  dj^-\-2xz=:  5. 

Suggestion.— Jlultiply  by  3  and  ^x'  +  Qx  =  15.  Whence  9a;'  +  6a; 
+  1  =  16.    a;=l,  and— f. 


356  QUADRATIC    EQUATION'S. 

49.  Solve  110ic2— 21a:  _  _i.   Result,  x  =  ^^  and  ^. 

Suggestion.— Multiply  by  110,  and  (IIO)V— 21  x  110a;  =  -110. 
Whence  llOV-31  x  110a;  +  (-V-)^  =  \,  and  llOx— ^  =  ±f 

50.  Solve  Z7?-\-6x  —  2, 

Suggestion.  (3)V  +  3  •  Sx  +  d)''  =  6  +  ^  =  \K  Whence  3a;  +  |. 
=  ±  |.     .-.  x  =  ^,  and  —2. 

Scholium. — It  appears  that  by  this  method  the  term  to  be  added 
to  complete  the  square  is  the  square  of  |  the  coefficient  of  the  first 
power  of  X.  Therefore  when  this  coefficient  is  odd,  fractions  arise. 
These  can  always  be  avoided  by  doubling  the  equation  when  this 
coefficient  is  odd,  before  completing  the  square. 

61.  Solve  Zofi-1x  =  ^0. 

Solution.— Multiplying  by  2  to  avoid  fractions,  Qx^—lix  =  80. 
(6)  V— 6  •  14CC  +  (7)*  =  49  +  480  =  529.     6aj-7  =  ±  23.     «  =  5,  and 

52.  Solve  ^x{x-\-b)  =  96  +  4a;(l— a;),  and  verify. 

Suggestion. — Perform  as  few  multiplications  as  possible.  When 
the  square  is  completed,  this  stands  (14)V  + 14  •  22a;  +  (ll)'=14  •  192 
+  121  =  2809. 

53.  Solve  ^x^^ix-{-^  =  0. 

54.  Solve  J-  _  _?_  =  |. 

a;— 2       xi-2       5 

55.  Solve  (ax—b)  (bx—a)  =  c^. 

Suggestion.  abx''—{a''  +  y)x  =  e—ab.  2a&a:'-(m)  =  2c'-2aft, 
letting  (m)  stand  for  the  term  which  becomes  the  middle  term  of  the 
trinomial  square  and  which  disappears  in  the  subsequent  process. 
Then    (2aZ>)-x-  -  (m)  +  {a"  +  &'0'  =  («'  +  5')'  +  ^oM  -  4:aW.     .'.  x 

-  ^J^Ll  \/(<^''-&')'  +  4g^' 
~''      "~"      2a5 


56.  Solve  «-V^^^^:z:g  ^  _^ .    Roots,  x  =  a,  and  -|a. 

a-j-V^ax—x^      a— a; 

Suggestion- — Clear  of  fractions  and  condense. 


EQUATIONS  SOLVED  AS  QUADRATICS.       '6oa 


KCTI0M    IIL 


EQUATIONS   OF   OTHER    DEGREES   WHICH     MAY    BE 
SOLVED   AS   QUADRATICS. 

107*  Prop.  1. — vdny  Pure  Equation  (i.  e.,  one  con- 
taining the  unknown  quantity  affected  with  hut  one 
exponent)  can  he  solved  in  a  manner  similar  to  a 
Pure  Quadratic. 

Demonstration. — In  any  such  equation  we  can  find  the  vahie  of 
the  unknown  quantity  affected  by  its  exponent,  as  if  it  were  a  simple 
equation.  If  then  tlie  unknown  quantity  is  affected  witli  a  positive 
integral  exponent  it  can  be  freed  of  it  by  evolution;  if  its  exponent 
be  a  positive  fraction  it  can  be  freed  of  it  by  extracting  the  root  indi- 
cated by  the  numerator  of  the  exponent,  and  involving  this  root  to 
the  power  indicated  by  the  denominator.  If  the  exponent  of  the 
unknown  quantity  is  negative  it  can  be  rendered  positive  by  multi- 
plying the  equation  by  it  with  a  numerically  equal  positive  exponent. 
q.  E.  D. 

EXAMPLES. 

1.  Solve  y-.-=-y. 

Suggestion,  y'  =  8.  .-.  y  =  2.  Why  not  put  the  ±  sign 
before  the  2  ? 

2.  Solve  ^8  =  -  ~^-s'  Result,  V  =  -^• 

3.  Solve  3a;i— 5  =  %x^.  Result^  x  =  125. 

4.  Solve  |-  4-  1  =  5  (^  -  l).  Result,  x  =  32. 

6.  Soke  12arl  4.  |  =  1  +  ^.  Result,  x  =  27. 

o      ^3       y 


S5S  EQUATIONS    OF    OTHER    DEGREES. 

6.  Given  { x  +  {iax  + 1 7«2)i  \h  —  ^h^ ^i  to  find  the  value 
of  X.  Result,  xz=z  16a. 


7.  Solve  SaizF  =  2ari;"  +  bb.         Result, 


-^a/O"- 


8.  Solve  a— 1  =  Ja;"      —a;"    . 


-=(Mr 


9.  Solve  a;t  :=  27.    Also  a;t  =  4.     Also  2/t  =  32. 

i?oo/5,  81,  ±32,  and  8. 

Query. — Why  the  ±  sign  in  one  case  and  not  in  the  others  ? 


108.  Prop.  2. — Any  equation  containing  one  un- 
known quantity  affected  with  only  two  different 
exponents,  one  of  ivhich  is  twice  the  other,  can  he 
solved  as  an  Affected  Quadratic. 

Demonstration. — Let  m  represent  any  number,  positive  or  nega- 
tive, integral  or  fractional ;  then  the  two  exponents  will  be  repre- 
sented by  m  and  2m ;  and  the  equation  can  be  reduced  to  the  form 
aj^'^-fl^af"  =  q.    Now  let  y—x'^^  whence  y'^=x^''\  whatever  m  may  be. 

Substituting,  we  have  y'^+py  =  q^  whence  y  =  — ^  "^  i/ T  +?•  ^^^ 
y  zizx"^]  hence  a?  =  |  —  |  ±  \/j  +  qT .    Q.  e.  d. 

EXAMPLES. 
Ex.  1.  Solve  3:z;3_^42:ct  =  3321. 

Solution. — Let  y  =  x^,  whence  y^  =  x^;  and  3y''  +  42y  =  3321. 
From  this  y  —  27,  and  —41.  Taking  the  first  value,  x^  =  27.  .*.  x 
=  9.  Taking  the  second  x^  =  —41.  .\x=  /v^ieSl.  We  there- 
fore find  that  x  =  9,  and  \/lQ&i. 


SOLVED    AS    QUADRATICS.  359 

2.  Solve  x^-h7x^  =  44.  x  =  ±8,  and  ±(-ll)i 
Query.— How  many  values  ?    Which  are  imaginary  ? 

3.  Solve  ix^-^x^  =  39.  x  =  729,  and  (^V- 

4.  Solve  3a:«  + 42:^-8  =  3321.  a;  =  3,  and  -\/U. 

8  17 

6.  Solve  ^+2  =  -^.  a;  :^  4^  and  i\^2. 

Scholium. — It  is  not  necessary  to  substitute  am.ther  letter  for  the 
unknown  quantity  as  given  in  such  examples.  Thus,  in  Ex.  3, 
doubling,  to  avoid  fractions,  Sx^  +  2a;«^  =  78.  Completing  the  square 
8'a5*  +  (m)  +  l  =  8  •  78+1  =  625.  Extracting  root,  8xi  +  1  =  ±25, 
X*  =  3,  and  -^.     .:  x  =  729,  and  /^V- 

[Note. — Solve  the  next  six  without  substituting.] 

6.  Solve  a;io  +  31.r^  =32.  x  z=  1,  and  —2. 

7.  Solve  a;" +  13/"  =  14.  x=h  and  (-14)»". 

8.  Solve  32:^ -4a;'  =  7.  x  =  Q^  and  1. 

9.  Solve  3a;  +  2v^  =  l. 

10.  Solve  V2x—7x  =  —52.  a;  =  8,  and  ^. 

11.  Solve  x-^2Vax-\-c=0. 

X—  \^Va±Va—c}\ 

12.  Solve  x\/  — X  =  — -_— 

a;=  iVTiiA/I. 


360  EQUATIONS    OF    OTHER    DEGREES. 


Suggestion. — The  first  member  may  be  written  - — — — .  Hence 
dropping  the  denominator,  \/«»  ^^^  squaring,  Qx^—x*=:1-\-2x*+sbi*. 


109*  Prop.  3. — Equations  may  frequently  he  put 
in  the  form  of  a  quadratic  by  a  judicious  grouping 
of  terms  containing  the  unknown  quantity,  so  that 
one  group  shall  he  the  square  root  of  the  other. 

Demonstration. — This  proposition  will  be  established  by  a  few 
examples,  as  it  is  not  a  general  truth,  but  only  points  out  a  special 
method. 

EXAMPLES. 


Ex.  1.  Solve  2ai2  +  3a;— 5V2a;2_^3a;4.94_3  =  o. 

Solution. — Add  6  to  each  member  and  arrange  thus,  (2a;^  +  3a:  +  9) 

— 5(2x''  +  3x  +  9)^  =  6.  Put  (2x'^  +  3a;4-9)i  =  y.  and  the  equation 
becomes  y'^-5p  =  G.  Whence  y  =  6,  and  —1.  Taking  y  =  6, 
2a;''4-3aj+9  =  36.     Whence  x  =  S,   and    —4^.      Taking  y  =  —  1, 

235'^  +  3aj+ 9  =  1.     Whence  x  =  ~^^Y~^J . 

4 

2.  Given  (2a; +  6)^(20;  + 6)^  =  6,  to  find  the  values 
of  X, 

Suggestion. — Put  y  =  (2x-f-6)^ ;  whence  y^  +  y  =  Q,  y  =  2,  and 
-3.     .'.  2x  +  Q  =  16,  and  also  2.r  +  6  =  81.     x  =  5,  and  37f 

Query.— Will  the  value  x  =  37^  verify  ?     Why  ? 

Jns. — Since  (2jr  +  6)^  and  (2«4-6)4  Hre'eveji  roots,  their  signs  are 
strictly  ambiguous,  though  not  so  expressed  in  the  example.  Sub- 
stituting for  .r,  37j^,  the  equation  becomes  (81)^"4-(81)i  =  6.     If 

now  we  regard  (81)*  —  —3,  as  it  is,  as  really  as  it  is  -f  3,  the  value 
verifies.     Such  cases  arc  frequent. 

3.  Given  (.r  +  12)^  =--  C>  —  {x  +  12)^,  to  find  the  values  of 
X,  and  verify  both  values. 


SOLVED    AS    QCADRATICS.  J^fil 

112 

4.  Given  j^. -^^  =  5+71^ ru '  to  find  the  values  otx. 

(2a;— 4)*      8     (2a:— 4)^ 

Values,  X  =  3,  and  1. 
Suggestion.— Put  (2a;— 4)'  =  y. 

5.  Solve  a; +5-^/3^+ 5  =  6.  a;  =  4,  and— 1. 


6.  Solve  2V^-3a;  +  ll  =  a:2— 3a;4-8. 

Suggestion.  z^—Sx+Q—2  \/x''—Sx+li  =  0.  Add  3  to  each 
member  and  x^-Zx  +  ll—^^x^—dx+li  =  3.  Put  \^x'—dx  +  li 
=-.  y,  and  y'— 2y  =  3.     .-.  .i-  =  2,  1,  and  i(3  ±  ^-31.) 

7.  Solve  (a:3-9)2=:3  +  ll(a:2_2). 

X  =  ±5,  and  ±2. 

8.  Solve  (x-^^\x  =  ^2-^' 

a;  =  4,  2,  andi(-7±A/T7). 

9.  Solve  x^f^l  +  ^^-idx^-^x)  =  70. 

Suggestion.  x*{l-\-—y  =  ^(Zx''+xy.  Hence  the  equation  may 
be  written  (3a;'  +  x)'— 9(3«'+a;)  =  630. 

Results,  X  —  3,  —31,  and  |(_1±  V-^1). 

10.  Solve  ^         -  =     "7   •  i?oo^5,  a:  =  4,  and  1. 

x—vx  ^ 

Suggestion.— Divide  by  «+  /y/ir,  and ^  =  — j^  .     Hence 

4  =  (x—  ■^^xy,  or  ar—  <v/x  =  ±  2. 

i/0.  Cor. — The  form  of  the  compound  term  may 
smnetimes  be  found  by  transposing  all  the  terms  to  the  first 
member,  arranging  them  with  reference  to  the  unknmvn  quan- 
tity, and  extracting  the  fiquare  root. 

In  trying  this  expedient,  if  the  highest  exponent  is  not  even  it 
must  be  made  so  by  multiplying  the  equation  by  the  unknown 
16 


36-3         EQUATIONS  OF  OTHER  DEGREES. 

quantity.  In  like  manner  the  coefficient  of  this  term  is  to  be  made 
a  perfect  square.  When  the  process  of  extracting  the  root  terminates, 
if  the  root  found  is  a  part,  or  a  factor,  or  a  factor  of  a  part  of  the 
remainder,  the  root  may  be  made  the  compound  term. 

11.  Solve^-^+-^i-  =  ^^  +  4. 

Solution. — Cleared  of  fractions,  transposed  and  arranged,  this 
becomes  dx*~4:2x^  +  lQSx''—U'7x—lS0  =  0.  Dividing  by  3,  x*—Ux^ 
+  56a;^-49aj--60  =  0. 

Extracting  square  root,  x* — lix^  +  560;'^ —49a!— 60 1  a;'^ — 7a;. 

X' 


^-7x\-Ux'  +  5Qx'' 
-14a;^  +  49a;' 

7a!'-49a;-60 
7(a;'-7a;)-60 

After  obtaining  two  terms  of  the  root  we  observe  the  root  itself  as 
a  factor  in  part  of  the  remainder.  The  equation  may  therefore  be 
written  (a;'-7a?)'  +  7  (x''-7x)—G0  =  0.  This  is  solved  as  before. 
X  =4,  3,  and  ^(7  ±  ^^69). 

12.  Solve  x^—12a^  +  ^4:X^-4:8x  =  9009. 

One  value  ofx  is  13. 

13.  Solve  7?-Qx^-\-llx—(j  =  0,  a;  =  1,  2,  3. 

14.  Solve  4a^ + f  =  4a?3 + 33. 

ar  =  2,  —I,  and  ^  (1  ±  V— 43). 

15.  Solve  7^-2c(^+x  =  a.  x  =  i±^f  ±  V«Ti. 

16.  Solve  a;* +a;3—4a:2+-j;  4.1=^0. 

a?  =  1,  and ^ 

Solution.- -Dividing  by  .t',  aj^  +  a?— 4  +  -  +  -  =  0,  which  may  be 

3/  X 

written  rc^  +  24--,+a;  +  -  =6,  or  ix  +  ^    +  ix  +  -\  =  6.      Whence 

X^  X  \       ^/  \       V 

the  compound  term  appears. 


SOLVED    AS    QUADRATICS.  363 

111,  Prop.  4.—  When  an  equation  is  reduced  to  the 

form  X"  +  Ax"-i  +  Bx'»-2 -f- Cx»»-« \-L  =  0,  the  roots. 

ivith  their  signs  changed,  are  factors  of  the  absolute 
(known)  term,  L. 

Demonstration.— Ist  Tlie  equation  being  in  this  form,  if  a  is  a 
root,  the  equation  is  divisible  by  x—a.  For,  suppose  upon  trial 
x—a  goes  into  the  polynomial  x*  +  Aaj"~' -f ,  etc.,  Q  times  with  a  re- 
mainder R.  (Q  represents  any  series  of  terms  which  may  arise  from 
such  a  division,  and  R,  any  remainder.)  Now,  since  the  quotient 
multiplied  by  the  divisor,  +  the  remainder  equals  the  dividend, 

we  have  {x—a)  Q  +  R  =  aj« + Aa;"-'  +  Bx"-''  +  Caj"-^  + +  L.     But 

this  polynomial  =  0.  Hence  (x-a)Q  +  R  =  0.  Now%  by  hypothe- 
sis a  is  a  root,  and  consequently  x—a  =  0.  Whence  R  =  0,  or  there 
is  no  remainder. 

2nd.  If  now  x—a  exactly  divides  «"  +  Aaj^'  +  Ba?^« + Caf-*-  -  +  L, 
a  must  exactly  divide  L,  as  readily  appears  from  considering  the 
process  of  division.  Hence  —a  is  a  factor  of  L,  a  being  a  root  of 
the  equation,     q.  e.  d. 


EXAMPLES. 
Ex.  1.  Solve  2a:*-C8a;  =  32-17^:3. 

Solution. — This  may   be  written  «*  +  — ^ — 34x— 16  =  0.     By 

« 

Prop.  4  the  roots  of  this  equation  are  factors  of  16.  We  therefore 
try  in  order  ±1,  ±2,  ±4,  ±8,  ±16  (all  the  integral  factors  of  16)  till 
we  find  whether  there  is  an  integral  root.     We  see  at  once  that 

neither  +1  nor  —1  satisfies  the  equation.    Trying  +2  we  find  it  is 

17a;' 
a  root.     Hence  x—2  is  the   divisor  sought.     Dividing  x*+——- 

2 

21 
— 34aj— 16  =  0,  by  .r— 2,  we  have  x*+  —  x'-\-21x  +  8=:0.     But  as 

2 

—2  satisfies  the  equation  as  well  as  +2,  x  +  2  is  a  divisor.   Dividing 


364        EQUATIOKS  OP  OTHER  DEGREES. 

21  17 

x^-\--—x^  +  21x-\-S  =  0  by  x  +  2,  we  have  x'  +  —  a;  +  4  =  0.    vFrom 
Z  * 

this  equation  ac  =  —  8,  and  —  ^.     The  roots  therefore  are  2,  —2, 

2.  Solve  a;-l  =  2  +  -4=- 

Suggestion. — Put  y  =  -y/o;,  and  clearing  of  fractions  and  trans- 
posing, y^—St/~2  =  0.  If  there  are  integral  roots  of  this  equation 
they  are  ±1,  or  ±2.  +1  does  not  satisfy  the  equation  and  hence  is 
not  a  root.  But  —1  does.  Hence  y  +  1  is  a  divisor.  Dividing,  we 
get  y^—y—2  —  0.  Whence  y  =  2,  and  —1.  .-.  The  roots  of  the 
equation  y^—dy—2  =  0  are  —1,  2,  and  —1,  there  being  two  equal 

roots.    Now  as  y  =  ^/x,  «  =  1,  and  4. 

3.  Solve  a^^dx^-^x-\-2  =  0. 

T7ie  irrational  roots  are  ^(l  ±  Vs). 

4.  Solve  a^  =  6x-^9. 

Imaginary  roots,  i(  ~  3  ±  V — 3). 

5.  Solve  a^—6x^—x-^S0  =  0. 

6.  Solve  x^-lW+6dx-\-'77  =  0. 

7.  Solve  x-^-Hxi  =  22.    (Put  x^  =  y.) 

Imaginary  roots,  29T7\/— 10. 

8.  SolYex*—2x^—l3x^—4:X—30  =  0. 

12  +  8^^ 


9.  Solve  X 


x—6 

Imaginary  roots,  i(— 3=Fa/— 7). 


10.  Solve  j:3— 2^72  +  457  +  7  =  0. 

11.  Solve  (x^+y  =  7,  and  y^  +  x  =  11. 


SOLVED    AS    QUADRATICS.  365 

Suggestions. — From  the  Ist,  x*—'7—p,  and  from  the  2d.  a:''=121 
— 22y'4-y'.  Whence  y*— 22y'-hy  +  114  =  0.  From  this  we  readily 
find  y  =  3 ;  whence  from  either  of  the  given  equations  we  can  find  a 
value  of  X. 

There  are,  however,  three  other  values  of  each  of  the  unknown 
quantities.  These  can  not  be  found  without  some  knowledge  of 
Higher  Equations. 

This  example  is  inserted  here  because  it  is  so  often  propounded  to 
teachers.  It  is  not  a  proper  example  to  propose  for  solution  by  the 
lower  algebra,  i.  e.,  by  the  method  of  quadratics.     (See  113.) 

12.  Solve  x*^a^—2  =  0. 

Also    x^—2a^-j-x—2  =  0. 
Also    a^-i-6a^  +  llx-\-6  =  0. 

,^    ^  ,       18   ,   Sl-a^      ar»-65 
•  ^^-  ^^^^^  ^  + -9^  = -72- 

Suggestion. — Putting  this  in  the  form  required,  the  absolute  term 
is  —1296;  the  integral  factors  of  which  are  very  numerous.  But 
trying  ±1,  ±2,  ±3,  ±4,  ±6,  ±8,  ±9,  we  find  —4  and  9  to  be  two  of 
the  roots. 

The  method  of  solving  this  by  putting  it  in  form  so  as  to  com- 
plete the  square  of  both  members,  is  to  multiply  by  2,  and  add 

8«      81       ^  ,^  ^  ^,  .  .       86       18       9       «'       2x       4 

-  +- toboth  members,  obtammg-  +^^+^  =  -+-+-. 

Extracting  root,  --|--  =  ±j^-l--J.     Whence  a?  =  9,  —4,  —4,  and 
—9,  there  being  two  roots  —4. 

14.  Given  x—S  =  —^ to  find  the  roots. 

X 

Suggestion- — The  roots  being  all  surds  and  imaginaries  in  this 
equation  cannot  be  found  by  the  principle  in  the  proposition.  The 
following  special  expedient  will,  however,  effect  a  solution  : 

Clear  of  fractions  and  add  x  +  1  to  each  member  and  aj*— 2a;-f-l 

=  A  +  Ayx+x. 


366  EQUATIONS    SOLVED    AS    QUADRATICS. 

14.  Solve  the  following  by  the  principle  in  the  propo- 
sition : 

x—1  _3  15  x-\-l 
x-{-6  ~  X  X  x+b 
^  +  x3-4a:2+a;  +  l  =  0.     (See  Ex,  16,  Prop.  3.) 

[Note. — Many  of  the  examples  under  Prop.  3  can  be  readily 
solved  in  this  manner.] 


SIMULTANEOUS   EQUATIONS  OF  THE   SECOND   DE- 
GREE   BETWEEN   TWO    UNKNOWN    QUANTITIES. 

112,  Prop.  1. —  Two  equations  between  two  unknown 
quantities,  one  of  the  second  degree  and  the  other  of 
the  first,  may  always  he  solved  as  a  quadratic. 

Demonstration.— The  general  form  of  a  (Quadratic  Equatim 
between  two  unknown  quantities  is 

since  in  every  such  equation  all  the  terms  in  x^  can  be  collected  into 
one,  and  its  coefficient  represented  by  a ;  all  those  in  a^  can  also  be 
collected  into  one,  and  its  coeflScient  represented  by  5,  etc. 

The  general  form  of  an  equation  of  the  First  Degree  between  two 
unknown  quantities  is 

a'x^Vy^-d  =0. 

"hfli /♦' 

Now,  from  the  latter  x  = -, — ,  which  substituted  in  the  for- 

a' 

raer  gives  no  terra  containing  a  higher  power  ofy  than  the  second, 

and  hence  the  resulting  equation  is  a  quadratic,     q.  e.  d. 


SIMULTANEOUS    EQUATIONS.  367 


EXAMPLES. 


Ex.  1.  Given  x-"^  =  4,  and  y-"^  =  1. 

Suggestion.— From  the  first  x  =  8— y.     Substituting  this  value 

8  — y  +  3y      ^  8  +  2y 

of  a;  in  the  second,  we  have  y— ^ —tt  =  1,   or  y—j^ —  =  1 ; 

o  —  y  •\-  i  lu— y 

whence  y^—^y=  —18,  and  y  =  Q  and  3. 

2.  Given  x+y  =  11,  a.ndx^-^2y^  =  34.    Verify. 

3.  Given  -  +  -  =  2,  and  x-\-y  =  2.    Verify. 

X      y 

4.  Given  x+y  =  100,  and  xy  =  2400. 

1       1       14 
6.  Given  2a;+3y  =  37,  and  -  +  -  =  —. 
^  a:       y       45 

6.  Given  2a^-i-xy—6y^  =  20,  and  2a;— 3^^  =  1. 

7.  Given  x  +  i  =  ?^^  and  ^±i^  =  1^. 

*  3  a;  2 

8.  Given  .13^  +  .  125a;  =  y—x^  and  y  —  .bx  =  .76xy—3x. 

Results,  X  =  0,  and  4 ;  and  y  =  0,  and  5. 

9.  Given    .3a;H-.125y  =  3^7— y,    and    3x—.6y  =  2.25xy 
+  3y. 

Results,  X  =  0,  and  —1 ;  and^  =  0,  and  —  2|. 


lis.  Prop.  2.--I1V  general,  the  solution  of  two 
quadratics  between  two  unknown  quantities,  requires 
the  solution  of  a  biquadratic. 

Demonstration. — Two  General  Equations  between  two  unknown 
quantities  have  the  forms 

(1)  aj;'  +  &»y  +  cy'  +  <ir  +  ey4/=  0,  and 

(2)  a'x"'  +  b'xy  -I-  cY  +  dx-{-^y-\-f'  =  0. 

From   (1).  .  =  -'J^  .  /<WI-g±^±r. 


368  SIMULTANEOUS    QUADRATIC    EQUATIONS. 

Now,  to  substitute  this  value  of  x  in  equation  (2),  it  must  be 
squared,  and  also,  in  another  term,  multiplied  by  y,  either  of  which 
operations  produce  rational  terms  containing  y'^,  and  a  radical  of  the 
second  degree.  Then,  to  free  the  resulting  equation  of  radicals  will 
require  the  squurmg  of  terms  containing  y^^  which  will  give  terms 
in  y,  as  well  us  other  terms,     q.  e.  d. 

[Note. — Since  it  is  not  the  purpose  of  this  treatise  to  embrace  the 
resolution  of  the  higher  equations,  only  such  special  cases  of  Simulta- 
neous Quadratics  with  two  unknown  quantities,  will  be  introduced, 
as  can  be  resolved  by  the  methods  of  quadratics.] 


114:,  Prop.  3. — Two  Homogeneous  Quadratic  Equa- 
tions between  two  unknown  quantities  can  always  he 
solved  by  the  method  of  quadratics,  by  substituting 
for  one  of  the  unhnown  quantities  the  product  of  a 
new  unhnown  quantity  into  the  other. 

Definition. — A  Homogeneous  Equation  is  one  in  which  each  term 
contains  the  same  number  of  factors  of  the  unknown  quantities. 
^x^  —  %xy  —  2/"  =  16  is  homogeneous.  3«^  —  ^y  +  y"^  =  10  is  not 
homogeneous. 

Demonstration. — The  truth  of  this  prox)osition  will  be  more 
readily  apprehended  by  means  of  a  particular  example.  Taking 
the  two  homogeneous  equations  x^—xy-\-y^  =  21,  and  y'"'— 2xy-|-15 
=  0.  Let  X  =  vy^  v  being  a  new  unknown  quantity,  called  an 
auxiliary,  whose  value  is  to  be  determined.  Substituting  in  the  given 
equations,  we  have  v^y^—vy^  +  y^  =  21,  and  y'^—2vy^  =  —15.    From 

21  15 

these  we  find  y^  =  -^ -,  and  y^  = 5.  Equating  these  values 

v^—v  +  l  ''        2v—l 

21  15 

Df  y\ =  - — -  ;  whence  42«-21  =  ISc'^-lSiJ  +  lS.     This 

•^  '  ^a  —  D  +  i       2v— 1 

latter  equation  is  an  affected  quadratic,  which  solved  for  «,  gives 

«  =  3,  and  ^.     Knowing  the  values  of  v  we  readily  determine  those 

15  - 

of  y  from  y^  — -r ,  and  find  y  =  ±  \/3  when  «  =  3,  and  y  =  ±  5 

2t)  —  1 

when  V  =  f .     Finally  as  x  =  «y,  its  values  are  a?  =  ±  3-^/3,  and  ±  4. 


WITH    TWO    UNKNOWN    QUANTITIES.  369 

By  obserdng  the  subatitution  oivy  tor  x  in  this  solution  it  iaseen 
that  it  brings  the  square  of  y  in  every  term  containing  the  unknown 
quantities,  in  each  equation,  and  hence  enables  us  to  Hud  two  values 
of  y-  in  terms  of  v.  It  is  easy  to  see  that  this  will  be  the  case  in  any 
homogeneous  quadratic  with  two  unknown  quantities,  for  we  have  in 
fact,  in  the  first  of  the  given  equations,  all  the  variety  ot  terms  which 
such  an  equation  can  contain.  Again,  that  the  equation  in  v  will 
not  be  higher  than  the  second  degree  is  evident,  since  the  values  of 
y'  consist  of  known  quantities  for  numerators,  and  can  have  denom- 
inators of  only  the  second,  or  second  and  first  degrees  with  reference 
to  V.  Whence  v  can  always  be  determined  by  the  method  of  quad- 
ratics; and  being  determined,  the  value  of  y  is  obtained  from  &pure 

15 
quadratic  (y'  =  r r ,  in  this  case),  and  that  of  x  from   a  dimple 

equation  {x  =  vy  in  this  case). 

EXAMPLES. 

Ex.  1.  Given  3a^-\-xy  =  18,  and  4.f-{-3xy  =  54. 

Roots,  X  =  ±2,  and  ±2V^ ;  and  y  =  ±3,  and  ±3VS. 

2.  Given  x^-]-xy-\-2y^  =  74,  and  2x^-j-2xy-^y^  —  73. 

Roots,  X  z=z  ±3,  and  =F8;  and  y  =  ±6. 

3.  Given  a^-\-3xy  z=z  64,  and  xy-\-4.y^  =  115. 

Roots,  a:  =  ±3,  and  ±36  ;  and  y  =  ±5,  and  ±11^. 

4.  Given  2x^-\-3xy  =  26,  and  Sy^-j-2xy  =  39. 

Roots,  ic  =  ±2,  and  y  =  ±3.     The  other  roots  are  oo . 

6.  Given  x^--4:y^  —  9,  and  xy^2y'^  =  3. 

15  3 

Roots,  X  =  ±--=  ;  and  y  =  ±— =r-    The  other  roots 
V2i  ^  A/2i 

are  oo. 

6.  Given  d7^-\'Xy—9  =  Q,  and  4yH3xy— 4  =  50. 
(See  Ex.  1.)  Roots.  x=  ±2,  y=  ±3. 

7.  Given  x^—xy  =  70,  and  xy—y^  =  12. 

8.  Given  a^-]-xy  :=  84,  and  x^—y^  =  24, 


370  SIMULTANEOUS    QUADKATIC    EQUATIONS 

115.  Prop.  4. —  When  the  unhnowri  quantities  are 
similarly  involved  in  two  quadratic,  or  even  higher 
equations,  the  solution  can  often  he  effected  as  a  quad- 
ratic, by  substituting  for  one  of  the  uuhnown  quanti- 
ties the  sum  of  two  others,  ajzd  for  the  other  unknown 
quantity  the  difference  of  these  new  quantities. 

[Note. — As  this  and  the  following  are  merely  special  expedients^ 
they  need  no  demonstrations  other  than  is  furnished  by  applying 
them  to  examples.] 

EXAMPLES. 

Ex.  1.  Griven  x'^-\-y^  =  52,  and  x-\-y+xy  t=  34. 

Solution. — In  these  equations  x  and  y  are  similarly  involved,  and 
hence  I  try  the  expedient  of  putting  x=m-\-n,  and  y=m—n,  whence 
x'  +  y^  -  2m'  +  2n^  =  52,  and  x  +  y  +  xy  =  2m-\-  ni'—ii'  =  34.  Now, 
from  the  two  equations  m'^  +  n^  —  26,  and 

2m  +  m'— w'  =  34,  by  adding 

I  have  2m-h2m'^     —    60,  whence  I  find  m= 5,  and 

—6.  Substituting  these  values  in  m'  +  n'=26,  n  =  ±l,  and  ±  \/— 10. 
Whence  the  reed  values  of  a;  are  found  to  be  6,  and  4 ;  and  of  y,  4, 
and  6. 

2.  Given  x^-\-x+y  =  IS—y^,  and  xy  =  6. 

Rational  roots,  x  =  3,  and  2  ;  and  y  =  2,  and  3. 

3.  Given  -t-^  +  iEn^  =  15,  and  x^-^y^  =  45. 

x-y  ^  x  +  y        '6'  ^^ 

fti       1%        10 
Suggestion.-- Using  the  same  notation  as  above,  — 1 —  =  •«■ » 

and  2m'  +  2n'=45;  whence  3m'  +  3;i'  =  10m?^,  and  wehave47W7i=27, 

27  9  3  3  9 

or  m  =  --.    7n  =  ±  -  ,  and  ±  - ;  n=  ±  -  ,  and  ±  -  .     The  roots  are 

4:Ti  iii  a  a  Z 

a?  =  ±  6 :  and  j/  =  ±  3. 

4.  Given  ^x-\-y)  =  3xy,  and  x-\-y-^a^+y^  =  26. 
Roots,  X  =  4,   and   2  ;  y  =  ^,  and  4.     Also  a;  =  — ^ 

±i\/377,  and  y  z^  -J^iFiVsr?. 


WITH    TWO    UNKNOWN    QUANTITIES.  371 

Definition  and  Scholium. — It  will  be  observed  that  the  above 
equations  are  of  the  second  degree,  and  that  they  have  the  unknown 
quantities  similarly  involved  ;  that  is,  in  the  last,  lor  example,  in 
the  first  member  there  is  +4a;,  and  also  +4y;  in  the  second  mem- 
ber X  and  y  are  multiplied  together ;  in  the  second  equation  there  is 
-fa-,  also  4-y;  there  is  +^'',  and- also  -vif.  Such  equations  can 
usually  be  readily  solved  in  this  manner.     But  the  equations  .c^  -f-y' 

5  1 

=  -ary,  and  x^y  —  -^y   have   the    unknown    quantities    similarly 

involved  in  the  firat  but  dissimilarly  in  the  second.  There  is  +a? 
in  the  second,  but  no  +y,  hence  they  are  not  similarly  involved. 
Whether  the  solution  of  such  equations  -svill  be  facilitated  by  this 
expedient  can  be  ascertained  only  by  trial.  In  this  case  the  expedient 
will  be  found  successful. 

5.  Given  oc^-\-y^  =.  ^xy,  and  x—y  =  Ixy. 

Roots f  a;  =  0,  4,  and  —  2  ;  y  =  0,  2,  and  —4. 

6.  Given  x'^-\-xy-\-4:y^  =  6,  and  3x^-\-Sy^  =  14. 

Roots,  X  =  ±2,  and  ^iVlO ;  -dnd  y  =  ±1,  and  ±f  VlO; 

7.  Given  x^—^y^  =  9,  and  xy-\-2y^  =  3. 

Suggestion. — The  student  will  find  by  experiment  that  the  above 
expedient  is  of  no  service  in  this  example.  The  example  is  readily 
8t)lved  by  finding  the  value  of  x  from  the  first  equation  and  sub- 
stituting it  in  the  second,  thus  obtaining  y  '\/4y*  +  9  =  3  —  2y^   or 

iy*  +  9y'  =  9—12y''  +  4y*.  Whence  21y^  =  9,  and  y  =  ±  y  -.  Or 
the  equations  can  be  treated  as  in  (114),  they  being  homogeneous. 

8.  Given  -  -f  ^  =  18,  and  x-j-y  =  12. 

y  ^  X  ^^ 

Suggestion.— In  these  equations  the  unknown  quantities  are 
similarly  involved,  and  although  the  first  is  of  the  third  degree  the 
expedient  of  the  proposition  is  successful,  r  =  8,  and  4  ;  and  y=4, 
and  8. 

Scholium. — In  all  symmetrical  equations  the  value  of  the  unknown 
quantities  must,  of  course,  be  the  same  numerically,  but  taken  in 
the  reverse  order,  since  the  lettei-s  can  change  places  in  the  equation 
without  altering  the  equations.    When,  therefore,  in  such  equations 


372  SIMULTANEOUS    QUADRATIC     EQUATIOJSTS 

the  values  of  one  of  the  unknown  quantities  are  found  the  values  of 
the  other  are  known. 

9.  Given  x^—x^y^-^y^  =  19,  and  x—xy-\-y  =  4, 

Suggestion. — Putting  x=m  +  n^  and  y=m—n,  x'^  +  y^  =  2m^  +  2n*. 
&ndx'^y^-(m  +  ny(m—ny  —  {m  +  n){/n—n)  x {m  +  7i){/7i—7i)—{?n'—n'^y. 
Hence  2m^  +  2n^—{m^—n^f  =  ld.     From  the  second  equation,  w^ 

Boots,  x  =  i(9±  VTS)  ;  and  y  =  i(^^^  V^)- 

10.  Given  a;  +  2/  =  11,  and  a;3_|.^  .-  407. 

Roofs,  X  =  7,  and  4  ;  and  ^  =  4,  and  7, 

11.  Given  2:— «/ =  3,  and  a:*  +  !/*  =  641. 

Moots,  x^  6,  and  —2 ;  y  =  2,  and  —5. 


SPECIAL    EXPEDIENTS. 


116»  Many  equations  of  other  degrees  than  the  second, 
and  which  do  not  fall  under  the  preceding  cases,  may  still 
be  solved  as  quadratics  by  means  of  special  artifices.  For 
these  artifices  the  student  must  depend  upon  his  own  inge- 
nuity, after  having  studied  some  examples  as  specimens. 
These  methods  are  so  restricted  and  special  that  it  is  not 
expedient  to  classify  them ;  in  fact,  every  expert  algebraist 
is  constantly  developing  new  ones. 

Ex.  1.  Given  x^-\-y^  =  5,  and  x^-\-y^  =  13. 

Suggestion. — In  case  of  fractional  exponents,  it  will  usually  be 
found  expedient  for  the  learner  to  put  the  unknown  quantities  with 
the  lowest  exponents,  equal  to  the  first  powers  of  new  unknown 
quantities,  and  thus  make  the  exponents  integral.     Thus,  putting 

x-^  =  w,  and  y^  =-  n,  we  have  x^  —  m^^  and  yt  =  n\  Whence  the 
equations  become  m  +  n  =5,  and  m^  +  n^  z=  13.  These  equations  are 
readily  solved  by  methods  already  learned,  and  we  find  m  ■=  3,  and 


WITH     TWO     rXKNOWN     <)UANTITrKS.  3T8 

2,  and  n  =  2,  and  3.     Hence  x^=3,  gives  a;  —  81 ;  and  a;*  =  2,  gives 
X  =  16.     Also  y^  —  2,  gives  y  =  8  ;  and  y»  =  8,  gives  y  =  27. 

2.  Given  x^-^y^  =  6,  and  x^-\-y^  z=z  126. 

Roots,  X  =  625,  and  1,  and  y  =■!,  and  3125. 

3.  Given  x^y^  =  2^^,  and  Sx^—y^  =14. 

4.  Given  a:— y  =  V''C-\-Vy,  and  a?^— //^  =  37. 

Suggestion. — Observe  that  both  members  of  the  first  are  divisible 
by  \/.r  +  \/y,  giving  a^x—  ^y  =  1.     x  =  16,  and  9  ;  y  =  9,  and  16 

6.  Given  7?-\-x-\-y  ■=.  18— y^,  and  xy  =  6. 

Suggestion. — From  the  first,  by  adding  twice  the  second,  we  may 

'writea5'  +  2.ry  +  2/'  +  x4-y  —  30,  or(aj  +  2^)'  +  (x  +  y)  =  30.    .-.  x  +  y=5, 
and  —6. 

6.  Given  x^+y^  =  52,  and  x  +  y-\-xy  =  3^. 

Roofs,  cc  =  6,  4,  and  — 6±\/~10;   «/ =  4,  6,  and  —6 


7.  Given  ic^-f  ?/=^  +  4\/^H^  =  45,  and  x*-^i/*  =  337. 

Suggestion. — In  the  first,  put  ^x^  +  y^  =  ©,  whence  t>'  +  4c  =■  45; 
and  t>  =  5,  und  —9.     a*  =  8,  and  4  ;  y  =  4,  and  8. 

8.  Given  ^  +  2-  =  GJI,  and  a^-^y^z=  65. 
Suggestion. — In  the  first,  put  —  =  r,  whence  v^  +  2vz=  9f|. 

Real  and  rational  roots,  x  =z  4:,  y  =7. 

0.  Given  x^+2xy-^y^-^2x  =  120— 2^^,  and  xy—y*  =  8. 
Roots,  y  =  \y   4,    —3  — a/5,   and  — 3-|-a/6;  a;  =  9,  6, 
—9  +  ^5,  and  — 9— \/5. 


^K4:  SiMULTANEOLS    EQUATIONS. 

10.  Given  xY  =  180— 8a;^,  and  x  +  dy  =  11. 

Roots,  X  =  5,  and  6  ;  y  =  2,  and  -J. 

11.  Given  a;^  +  3.t + ^  =  73 — 2xy,  and  y^  ^-  3y  +  ic  =  44. 

Suggestion. — Add  the  two  equations  together,  and  proceed  as 
before.      «  =  4,    16,  and    -13^^58;  and  y  =  5,   -7,  and   —  1 

±  V58. 

12.  Given  xy-\-xy^  =  12,  and  x-{-xy^  =  18. 

12 
Suggestion. — From  the  first,  x  =  ~- ;  and  from  the  second* 

18  12  18         _.  .   . 

if  =  r- — ^.     .'.  -,i ^  =  z i.     Dividing  denominators  by  l  +  i/, 

1+y'  2/(1  +  2/)       1  +  2/ 

2  3 

and  nunierators  by  6,  we  have  -  = :    whence   2  — 2^/  +  2i/* 

y      1— y  +  2/^' 

=  3y.     Hence  x  =  2,  and  16  ;  and  y  =  2,  and  ^. 

13.  Given  x^-\-xy-^y'^  —  26,  and  a^-ficy+J/^  =  364. 

Suggestion. — From  the  first,  x'^  +  y"-  =  2Q—xy;  and  from  the  sec- 
ond, by  adding  x^y^  to  both  members  and  extracting  the  square  root, 
x^  +  y'^  =  '\/dQ4^  +  x^y'^.  Equating  these  values  ofx'^  +  y^,  and  squar- 
ing, we  have  GlQ—52xy  +  x'^y'^  =  Siii  +  xy'^,  whence  xy  =  6.  Squar- 
ing this  and  adding  it  to  the  second,  and  extracting  the  square 
root,  jc^-l-y'  =  20.  Also  subtracting  dx'^y^  =  108  from  the  second, 
and  extracting  the  square  root,  x'^—y^  =  16.  Whence  x^  =  18,  and 
2/'  =  3.  ^ 

Another  Solution. — Dividing  the  second  by  the  first,  we  have  x' 
—xy  +  y"  —  14.  Subtra(jting  this  result  from  the  first,  we  have  2xy 
=  12.     Whence  the  solution  proceeds  as  above. 

[Note. — Though  this  field  is  illimitable,  it  is  not  thought  necessary 
for  the  learner  to  pursue  special  methods  farther,  inasmuch  as  what 
is  given  will  enable  him  to  catch  the  spirit  of  such  solutions, 
and  no  writer  in  discussing  a  problem  involving  processes  even  as 
complex  as  some  given  above,  would  fail  to  give  hints  at  his  methods 
of  solution.] 


OF    THE    SECOXD     DKGREE.  875 


APPLICATIONS. 

[Note. — One  of  the  most  important  things  to  be  learned  from  the 
following  examples  is  several  devices  frequently  found  serviceable 
in  stating  a  problem,  which  make  the  equations  arising  more  simple 
and  easy  of  solution.  These  devices  are  of  special  necessity  in 
examples  involving  progressions.] 

Ex.  1.  What  number  is  that  which  being  divided  by  the 
product  of  its  two  digits,  the  quotient  is  2,  and  if  27  is 
added  to  it  the  digits  are  reversed  ? 

2.  There  are  three  numbers,  the  difference  of  whose  dif- 
ferences is  8  ;  their  sum  is  41 ;  and  the  sum  of  their  squares 
is  699.    What  are  the  numbers  ? 

Notation. — Let  x  be  the  second  number,  and  y  the  diflference 
between  the  second  and  first,  so  that  x—y  represents  the  first.  The 
first  equation  is  3.C+8  =  41,  and  the  second  (11— 2/)'  +  131  +  (19+y)' 
=  699. 

3.  There  are  three  numbers,  the  difference  of  whose  dif- 
ferences is  5 ;  their  sum  is  44 ;  and  their  product  is  1950. 
What  are  the  numbers  ? 

4.  A  grocer  sold  80  lbs.  of  mace  and  100  lbs.  of  cloves  for 
165 ;  but  he  sold  GO  lbs.  more  of  cloves  for  $20  than  he  did 
of  mace  for  $10.     What  was  the  price  of  a  pound  of  each  ? 

5.  A  and  B  have  each  a  small  field,  in  the  shape  of  an 
exact  square,  and  it  requires  200  rods  of  fence  to  enclose 
both.  The  contents  of  these  fields  are  1300  square  rods. 
What  is  the  value  of  each,  at  $2. 25  per  square  rod  ? 

Ans.,  One,  $900;  other,  $2,025. 


6.  Find  two  numbers,  such  that  the  sum  of  their  squares 
being  subtracted  from  three  times  their  product,  11  remain  ; 
and  the  difference  of  their  squares  being  subtracted  from 
twice  their  product,  the  remainder  is  14.     (See  114.) 


376  SIMULTANEOUS    EQUATIONS 

7.  "What  two  numbers  are  those  whose  difference  multi- 
plied by  the  difference  of  their  squares  is  32,  and  whose  sum 
multiplied  by  the  sum  of  their  squares  is  272  ? 

8.  The  difference  of  two  numbers  is  2,  and  the  square  of 
their  quotient  added  to  four  times  their  quotient  is  9^. 
What  are  the  numbers  ?  Ans.,  o  and  3. 

9.  There  are  two  numbers,  whose  sum  multiplied  by  the 
less,  is  equal  to  four  times  the  greater,  but  whose  sum  mul- 
tiplied by  the  greater  is  equal  to  9  times  the  less.  What  are 
the  numbers  ? 

10.  Find  two  numbers,  such  that  their  product  added 
to  their  sum  shall  be  47,  and  their  sum  taken  from  the  sum 
of  their  squares  shall  leave  62.  A7is.,  5  and  7. 

11.  Find  two  numbers,  such  that  their  sum,  their  pro- 
duct and  the  difference  of  their  squares  shall  be  all  equal  to 
each  other.  Ans.,  f  ± jVS,  and  i±iVS. 

12.  Find  two  numbers  whose  product  is  equal  to  the  dif- 
ference of  their  squares,  and  the  sum  of  their  squares  equal 
to  the  difference  of  their  cubes. 

Ans.,  i\/5,  and  ^{5  +  \/5). 

13.  A  person  has  11,300,  which  he  divides  into  two  ])or- 
tions,  and  loans  at  different  rates  of  interest,  so  that  the 
two  portions  produce  equal  returns.  If  the  first  portion 
had  been  loaned  at  the  second  rate  of  interest,  it  would 
have  produced  136,  and  if  the  second  portion  had  been 
loaned  at  the  first  rate  of  interest,  it  would  have  produced 
$49.    Eequired  the  rates  of  interest. 

Ans.,  7  and  6  per  cent. 

14.  The  fore  wheel  of  a  wagon  makes  6  revolutions  more 
than  the  hind  wheel  in  going  120  yards;  but  if  the  periphery 
of  each  wheel  be  increased  1  yard,  the  fore  wheel  will  make 
only  4  revolutions  more  than  the  hind  wheel  in  going  the 


OF    THK     ."SECOND     DEGREE.  377 

same    distance.       What     i«    the    circiiniference    of    each 
wheel?  Ans.,  4  and  5. 

15.  The  sum  of  two  numbers  is  8  and  the  sum  of  their 

nubes  is  152  ;  what  are  the  numbers  ? 

Suggestion.— Let  x  +  y  be  one  of  the  numbers,  and  a;—y  the  other. 
Then  a;  =  4,  and  2y^-^Qxi/'  ^  152  (111). 

16.  The  sum  of  two  numbers  is  7,  and  the  sum  of  ihGir 
4th  powers  is  641.     What  are  the  numbers  ? 

17.  The  sum  of  two  numbers  is  G,  and  the  sum  of  their 
5th  power  is  1050.     What  are  the  numbers  ? 

18.  The  product  of  two  numbers  is  24,  and  their  sum  multi- 
plied by  their  difference  is  20 ;  find  them.    Aifs„  4  and  6. 

19.  What  two  numbers  are  those  whose  sum  multiplied 
by  the  greater  is  120,  and  whose  difference  multiplied  by 
the  less  is  16  ?  A7is.,  2  and  10. 

20.  What  two  numbers  are  those  whose  sum  added  to  the 
sum  of  their  squares  is  42,  and  whose  product  is  15  ? 

Ans.,  3  and  5. 

21.  A's  and  B's  shares  in  a  speculation  altogether  amount 
to  $500 ;  they  sell  out  at  par,  A  at  the  end  of  2  years,  B  of 
8,  and  each  receives  in  capital  and  profits  $297.  How  much 
did  each  embark  ?  A?is.,  A,  $275  ;  B.  $225. 

Suggestion. — Letting  j:  )}e  A's  capital,  and  y  B's,  A  gained  297 
—X,  and  B,  297— y.  And  as  the  gains  are  proportioned  to  the  pro- 
ducts of  the  respective  tiraes  into  the  capitals.  2x:8y  ::  297— a; 
:  297-y.  

22.  What  three  numbers  are  those  in  A.  P.,  whose  sum  is 
120,  and  the  sum  of  whose  squares  is  5600  ?  Ann.,  20,  40,  60. 

Suggestion. — In  solving  examples  involving  several  quantities  in 
arithmetical  progression,  it  is  usually  expedient  to  represent  the 
middle  one  of  the  series,  when  the  number  of  terms  \Hitdd,  by  .r,  and 
let  y  be  the  c^mon  difference.  If  the  number  of  terms  is  <?»e»,  rep- 
resent tlie  two  mi'We  f/'nm  hy  r— y,  and  x  +  y,  making  the  common 


378  i^IMULTANKOUS     KQUATIONS 

diflference  2</.  Thus  the  statement  of  the  above  problem  becomes 
x—y+x+x-\-i/^120,ordx  =  130;  &nd  (x-yY +x*  +  {x  +  yy=5Q00, 
or  3»'  +  22/'  =  5600. 

23.  What  four  numbers  are  those  in  A.  P.,  the  sum  of 
whose  squares  is  84,  and  their  product  105? 

Suggestion.— Call  the  numbers  x—dy,  x—y,  x-\-y,  and  x-\-%y. 

24.  The  sum  of  five  numbers  in  A.  P.  is  35,  and  the  sum 
of  their  squares  285  ;  find  the  numbers. 

25.  What  three  numbers  are  those  in  A.  P.,  the  sum  of 
whose  squares  is  1232,  and  the  square  of  the  mean  greater 
than  the  product  of  the  two  extremes,  by  16  ? 

26.  Find  four  numbers  in  A.  P.  such  that  the  sum  of  the 
squares  of  the  extremes  is  4500,  and  the  sum  of  the  squares 
of  the  means  is  4100. 

27.  The  product  of  five  numbers  in  A.  P.  is  945 ;  and  their 
sum  is  25.     What  are  the  numbers  ? 

Ans.,  1,  3,  5,  7,  9. 

28.  The  product  of  four  numbers  in  A.  P.  is  280,  and  the 
sum  of  their  squares  166  ;  find  them. 

29.  The  sum  of  nine  numbers  in  A.  P.  is  45,  and  the  sum 
of  their  squares  285  ;  find  them. 

Ans.,  1,  2,  3,  etc.,  to  9. 

30.  The  sum  of  seven  numbers  in  A.  P.  is  35,  and  the  sum 
of  their  cubes  1295 ;  find  them.         Ans.,  2,  3,  etc.,  to  8. 


31.  There  are  three  numbers  in  geometrical  progression, 
whose  sum  is  52,  and  the  sum  of  the  extremes  is  to  the 
mean  as  10  to  3.     What  are  the  numbers  ? 

Ans.,  4,  12,  and  36. 

Suggestion. — Let  x  be  the  first  term  and  y  thA  ratio.  Then 
x-\-xy-\-xy'^  —  52    and  x-\-xy'^ :  xy  : :  10  :  3,  or  1  +y^ :  y  : :  10  :  3. 


OF    THE    SKOOKD     DEGREE.  379 

32.  The  sum  of  three  numbers  in  geometrical  progression 
is  13,  and  the  product  of  the  mean  and  sum  of  the 
extremes  is  30.     What  are  the  numbers  ? 

Ans,,  1,  3,  and  9. 

Suggestion.— We  have  x+ry+x?/  ~13,  and  {x-{-xy^)xy  =  dO 
From   the   first  x  +  xy''  —  13— -ty,   and   from   the  secoad,    x-\-xy 

on  QQ 

=  -;  whence  13-2^=—,  or  jcy-13a:y= -30. 

33.  If  the  seventh  and  tenth  terms  of  a  geometrical  pro- 
gression are  6  and  750  respectively,  what  are  the  inter- 
mediate terms  ? 

Suggestion. — The  equations  are  ajy"  =  6,  and  xy*  =  750.  Divide 
the  second  by  the  first. 

34.  If  the  third  and  fifth  terms  of  a  geometrical  progres- 
sion be  75  and  300  respectively,  what  will  the  fourth  term 
be?  Ans.,  150. 

35.  If  the  first  and  fourth  terms  of  a  geometrical  progres- 
sion are  3  and  24  respectively,  what  are  the  two  inter- 
mediate terms*? 

36.  There  are  four  numbers  in  geometrical  progression. 
The  sum  of  the  means  is  30,  and  the  product  of  the 
extremes  200  ;   what  are  the  numbers  ? 

x^  y^  y 

Suggestion.— Represent  the  numbers  by  — ,  ar,  y,  and  ~,in  which  - 

is  the  ratio.     The  equations  are  x  +  y  —  30,  and  xy  =  200. 

x'  y' 

For  an  odd  number  of  terms,  as  five,  use  —  ,  a;',  xy,  y',  —  • 

y  X 

37.  The  sum  of  three  numbers  in  G.  P.  is  26,  and  the 
sum  of  their  squares  is  364  ;  required  the  numbers. 

Suggestion. — The  equations  are  x'  +  a^  +  y'  =  26,  and  ar*+a:'y'-f-y* 
=  364,  which  have  already  been  solved. 


380 


SYNOPSIS. 


CO 

z 
o 

a 


o 

Q 
< 


r  DEFINITIONS. 
PURE. 

AFFECTED. 


EQUATIONS 

OF  OTHER 

DEGREES. 


SIMULTA- 
NEOUS 
QUADRATICS. 


EXPEDIENTS 
^   IN  STATING. 


SYNOPSIS. 

Quadratic  Equation. -S^""-'"^"^^^*^ 
„      ^     ^  ( Affected.— Complete. 

Root  of  Equation. 

Di./xk      «,    o,  (  Cor.  1.  Number  of  Roots. 

rrOD.     To  Solve.    Dem.J  Dem. 

{  Cor.  2.  Imaginary  Roots. 

Prob.     To  solve.    RULB.    Dem. 

n^^^t^i,' o  (  Common  method. 

Completing  Square.     what?-J 

( Special  methods. 

Cor.  I.     Roots  of  an  Affected  Quadratic.    Dem. 
.  Cop.  2.     To  write  the  root  of  x^  +px  =  q  directly. 
f  Pure.     Prop.  1. 

iProp.  2.    Dem.    When  solved  as  Quad. 
I  P)vp.  3.    Cor. 
Expedients.  < 

(  Prop.  4.    How  applied. 


Affected. 


Prop.  I.  Dem. 
Prop.  2.  Dem. 
Prop.  3.     Dem. 

Prop.  4.     Sch. 

Expedients.     Enumerate  the  7  ^ven. 


An  A.  P. 


i  Common  method. 
Special  methods.  ■< 


Even  number  of  terms. 
Odd         "       "      " 


A  G.  P 


j  Common  method. 
( Special  methods 


{Even  number  of  terms. 
Odd         "       "     « 


Test  Questions. — If  one  of  two  equations  between  two  unknown 
quantities  is  of  the  first  degree,  and  the  other  of  the  second,  what 
will  be  the  degree  of  the  resulting  equation  after  eliminatinsr  one  of 
the  unknown  quantities?  Prove  it.  How  may  such  equations  lie 
solved  ?  ■  In  general,  what  is  the  degree  of  the  equation  arising  from 
eliminating  one  unknown  quantity  from  two  equations,  each  of  the 
second  degree  ?  Prove  it.  Mention  the  several  cases  given  in  which 
such  equations  can  be  solved  by  quadratics,  and  state  the  process 
in  each  case. 


ClAPTEl^.  Y. 


[Note. — It  Is  the  purpose  of  this  chapter  to  give  a  simple, 
arithmetical  view  of  the  nature  of  logarithms,  with  some  illustrations 
of  their  practical  utility.  For  the  production  of  the  Logarithmic 
Series  and  its  use  in  computing  Logarithms,  see  Appendix  IL 
Enough,  however,  is  here  given  for  practical  use  in  trigonometry,  as 
usually  studied,  and  to  enable  the  student  to  understand  the  use  of 
logarithms  in  ordinary  operations.) 

117.  A,  Logarithm  is  the  exponent  by  which  a  fixed 
number  is  to  be  affected  in  order  to  produce  any  required 
number.  The  fixed  number  is  called  the  Base  of  the 
System. 

Illustration. — Let  the  i^rtse  be  3  :  then  the  logarithm  of  9  is  2;  of  27, 
8;of81,4;  ofl9683,  9;  for  3'  =  9;  3''=27 ;  3^=81;  and8'=19683. 
Again,  if  64  is  the  base,  the  logarithm  of  8  is  ^,  or  .5,  since  648,  or 
64  *  =  8  ;  I.  e.,  |,  or  .5  is  the  exponent  by  which  64,  the  base,  is  to  be 
affected  in  order  to  produce  the  number  8.  So  also,  64  being  the 
base,  |,  or  .333+  is  the  logarithm  of  4,  since  64*,  or  64 •''^"  =  4; 
i.  e.,  I,  or  .333+  is  the  exponent  by  which  64,  the  base,  is  to  be 
affected  in  order  to  produce  the  number  4.  Once  more,  since  64», 
or  64  « «"  =  16,  f,  or  .666+  is  the  logarithm  of  16,  if  the  base  is  64. 
Finally,  64~i  or  64— *  =  |,  or  .125;  hence  — |,  or  —.5,  is  the 
logarithm  of  ^,  or  .125,  when  the  base  is  64.  In  like  manner,  with 
the  same  base,  — |,  or  —.333+  is  the  logarithm  of  |,  or  .25. 


EXAMPLES. 

Ex.  1.  If  2  is  the  base,  what  is  the  logarithm  of  4  ?  of  8  ? 
of  32?  of  128?  of  1024? 


382  LOGARITHMS. 

Model  Solution. 

7  is  the  logarithm  of  128,  if  2  is  the  base,  since  7  is  the  exponent 
by  which  2  is  to  be  aflfected  in  order  to  produce  the  number  128. 

2.  If  5  is  the  base,  what  is  the  logarithm  of  625  ?  of 
15625  ?  of  125  ?  of  25  ? 

3.  If  10  is  the  base,  what  is  the  logarithm  of  100  ?  of  1,000  ? 
of  10,000  ?  of  10,000,000  ? 

4.  If  2  is  the  base,  what  is  the  logarithm  of  J,  or  .25  ?  of 
i,  or  .125  ?  of  -^s,  or  .03125  ?  Ans.  to  the  last,  —5. 

5.  If  8  is  the  base,  of  what  numher  is  f,  or  .666-f-  the 
logarithm?  of  what  number  is  f  or  1.333+  the  logarithm? 
of  what  number  is  2  the  logarithm  ?  of  what  number  is  2^, 
or  2.333+  ?  of  what  number  3|,  or  3.666  +  ? 

Ans.  to  the  last,  2048. 
Scholium. — Since  any  number  with  0  for  its  exponent  is  1,  th6 
logaritlim  of  1  is  0,  in  all  systems.     Thus  10"  =  1,  whence  0  is  the 
logarithm  of  1,  in  a  system  in  which  the  base  is  10. 

118.  A  System  of  Logarithms  is  a  scheme  by  which 
all  numbers  can  be  represented,  either  exactly  or  approxi- 
mately, by  exponents  by  which  a  fixed  number  (the  base)  can 
be  affected.     Negative  numbers  can  have  no  logarithms. 

119.  There  are  Two  Systems  of  Logarithms  in  common 
use,  called,  respectively,  the  Briggea?i  or  Common  System, 
and  the  Napierian  or  Hyperbolic  System.  The  base  of  the 
former  is  10,  and  of  the  latter  2. 71828 +  . 


120.  One  of  the  most  important  uses  of  logarithms  is 
to  facilitate  the  multiplication,  division,  involution,  and  the 
extraction  of  roots  of  large  numbers.  These  processes  are 
performed  upon  the  following  principles  : 


121,  Prop.  l.  —  TIie  sum  of  the  logarithms  of  two 
numbers  is  the  logarithm  of  their  proditct. 


LOGARITHMS.  383 

Demonstration. — Let  a  be  the  baae  or"  the  systeiu.  Let  m  and  n 
be  any  two  nuQibers  whose  logarithms  are  x  and  y  respectively. 
Then  by  definition  a'  =  7n,  and  a^  =  n.  Multiplying  these  equations 
together  we  have  o'+i' =  mn.  Whence  x  +  y  \%  the  logarithm  of 
mil.      Q.  E.  D. 

122.  Prop.  2. — The  logarithm  of  the  quotient  of 
two  numbers  is  the  logarithm  of  the  dividend  minus 
th3  logarithm  of  the  divisor. 

Demonstration- — Let  a  be  the  base  of  the  system,  and  m  and  n 
any  two  numbers  whose  logarithms  are,  respectively,  x  and  y. 
Then  by  definition  we  have  a'  =  m,  and  (i^  —  n.    Dividing,  we  have 

tf*-"  =  — .    Whence  x—y  is  the  logarithm  of  — .    q.  e.  d. 


123,  Prop.  3. — The  logarithm  of  a  power  of  a 
number  is  the  logarithm  of  the  number  multiplied  by 
the  index  of  the  power. 

Demonstration. — Let  a  be  the  base,  and  x  the  logarithm  of  m. 
Then  a'  ==m]  and  raising  both  to  any  power,  as  the  eth,  we  have 
a~  =  m'.     Whence  0:2!  is  the  logarithm  of  the  «th  power  of  w.  q.  e.  d. 


124,  Prop.  4. — The  logai^ithm  of  any  root  of  a 
nuniber  is  the  logarithm  of  the  number  divided  by 
the  number  expressing  the  degree  of  the  root. 

Demonstration. — Let  a  be  the  base,  and  x  the  logarithm  of  m. 
Then  a?  —  m.      Extracting    the    eth   root    we    have    a*  —  A^m. 


X 


Whence  -  is  the  logarithm  oi\/m.     q.  e.  d. 


125,  In  order  to  apply  these  principles  practically,  we 
need  what  is  called  a  Table  of  Logarithms.  That  is,  a  table 
from  which  we  can  readily  obtain  the  logarithm  of  any 
number,  or  the  nnmbej-  corresponding  to  any  logarithm. 


384  LOGARITHMS. 

We  will,  fcherelore,  proceed  to  show  how  such  a  table  can 
be  computed.  Though  the  method  about  to  be  given  is 
not  the  most  expeditious  now  known,  it  is,  nevertheless,  the 
one  used  when  our  tables  were  first  computed. 


120,  Prob.— To  compute  the  common  logarithm  of  any 
decimal  number. 

Demonstration. — 1st.  It  is  evident  that  it  is  necessary  to  com- 
pute the  logarithms  oi'  jjnme  numbers  only,  since  the  logarithm  of  a 
composite  number  is  the  sum  of  the  logarithms  of  its  factors  (121). 
2n(l.  To  compute  the  logarithms  of  the  series  of  prime  numbers. 
In  the  first  place  we  know  that  the  logarithm  of  1  is  0,  since  10"  =  I, 
Also  the  logarithm  of  10  is  1,  since  10'  =  10.  Now  if  we  find  the 
logarithm  of  5,  we  can  get  the  logarithm  of  2  by  subtracting  the 
logarithm  of  5  from  log.  10.*  If  there  be  any  number  which  is  the 
logarithm  of  5  it  is  evident  it  must  lie  between  0.  which  is  log,  1, 
and  1,  which  is  log.  10.     Therefore  starting  with 

10"  =  1     and 

10^  =  10  multiplying  them  together 
we  have  10'  =  10 

Extracting  the  square  root,  10*"  —  y'lO  =  3.162377 +  . 
Again,  as  5  lies  between  10  and  3. 162277  +  its  logarithm  lies  between  1 
and  .5.  Multiplying  the  last  two  equations  we  have  10^-^=31.62277  +  . 

Extracting  the  square  root,  10-''   =  V^l  62277+  =  5.623413  + . 

Again,  10'*     =3.162277  + 

and  10-'^   =5.623413  + 

Multiplying,  10'-"  =  17.7827895914  + 

Extracting  the  square  root,  10-«"  =  ^17^7827895914  +  =4.216964  + . 

Again,  multiplying  this  last  by  10'"  =  5.623413 +  ,  as  5  lies 
between  these  numbers,  and  extracting  the  square  root,  we  have 
10-^^'^ =4.869674  .  In  each  case  the  exponent  of  10  is  the  logarithm 
of  the  number;  thus  .6875  is  log.  4.869674  +  .  Continuinsf  this 
process  to  22  operations  (!)  we  have  log.  5.000000+  =  .698970  +  , 
which  is  sufficiently  accurate  for  ordinary  purposes. 

Now  log.  10 -log.  5  =  log.  2  =  1 -.698970  =  .301030. 

*  This  is  the  common  abbreviation  of  "  logarithm  of  10,"  and  6Uould  be  read 
•* logarithm  of  10."  not  "  log  ten,"  which  is  grossly  inelegant. 


LOtiAltlTHMS.  385 

To  find  log.  8,  we  would  take  10'  =  3.162277  +  ,  and  10°  =  1,  and 
proceed  as  before. 

Were  it  our  purpose  to  find  log.  11,  the  computation  would  be  as 
follows : 

10*  =  10 

10"  =  100 

lO''  =  1000 

10   =10'-^=  ^1000  =  81.62277+ 

10'  =  10 


10l  =  10'-' =  316.2277  + 

10»-"  = 

V316.2277+  = 

17.78278  + 

10* 

= 

10 

10*-"  = 

:  177.8278  + 

lO'-'^'  = 

:  V177-8278+  = 

=  13.33521  + 

10* 

= 

10 

lO-'"  = 

=  133.3521  + 

10*-""  : 

=  V133.3521  +  : 

=  11.54782- 

10» 

= 

10 

lO'-""  : 

=  115.4782  + 

|Q1.0>13& 

=  Vl  15-4782  + 

=  10.74607  + 

|0t.o««  . 

= 

11.54782- 

|Q9.0»37b 

=  124.09368  + 

ID'-""""  =  Vi'-^-ooaes 

+  =  11.13973  + 

]^QI.U«13b 

= 

10.74607  + 

lO^.o^.m^  119.70845  + 
101.0890636  _  10.94113  + 

Whence  1.0390625  is  log  10.94113;  and  proceeding  with  the  com- 
putation, the  logarithm  of  11  maybe  found  with  sufficient  accuracy. 

Scholium. — The  pupil  will  not  fail  to  be  impressed  with  an  idea 
of  the  inmien.se  labor  involved  in  computing  a  table  of  logarithms. 
The  common  tables  give  the  logarithms  of  numbers  from  1  to  10,000, 
with  provision,  as  will  l)e  seen  hereafter,  for  using  {hem  to  find  the 
logarithms  of  much  larger  numbers,  with  sufficient  accuracy  for 
practical  purposes.     One  page  of  such  a  table  is  given.   (Page  386.) 


/?7.  Prob.— To  And  the  logarithm  of  a  number  from 
the  table. 

17 


386 

(a  page  of)  a 

TABLE  OF 

LOGARITHMS. 

N. 

0 

1 

2 

3 

4 

5 

6 

7  1 

1 

8 

9 

D. 

280 

447158 

7313 

7468 

7628 

7778 

7933 

8088 

82421 

8397 

8552 

155 

281 

8706 

8861 

9015 

9170  9324 

9478 

9633 

9787' 

9941 

.95 

154 

283 

450249 1 

0403 

0557 

0711108651 

1018 

1172 

1326 

1479 

1633 

154 

283 

1786! 

1940 

2093 

2247 

2400 

2553 

2706 

2859 

3012 

3165 

153 

284 

3318 

3471 

3624  3777 

S930 

4082 

4235 

4387 

4540 

4692 

153 

285 

4845 

4997 

5150  5302:5454 

5606  i  5758 

5910 

6062 

6214 

152 

286 

6366 

6518 

6670  6821  !  6973 

7125  7276 

7428 

7579 

7731 

152 

287 

7882 

8033 

8184  8336  8487 

8638  \  8789 

8940  9091 

9242 

151 

288 

9392 

9543 

9694 

9845 

9995 

•  146  1  .296 

.447 

•597 

.748 

151 

289 

460398 

1048 

1198 

1348 

1499 

1649li;99 

1948 

2098 

2248 

150 

290 
291 

2398 
3893 

2548 
4042 

2697 

2847 
4340 

2997 
4490 

3146  3296 

46S9  :  4788 

3445  3594 
4916  r>085 

3744 

5284 

150 
149 

4191 

292 

5883 

5532 

56-0 

5829 

5977 

6126  !  6274 

6423  6571 

6719 

149 

293 

6868 

7016 

7164 

7312 

7460 

7608 

7756 

7904  8052 

8210 

148 

294 

8347 

8495 

8643 

8790 

8938 

9085 

9233 

9880  9527 

£675 

148 

295 

9822 

9969 

.116 

•263 

.410 

.557 

.704 

.851  -998 

1145 

147 

29G 

471292 

1438 

1585 

1732 

1878 

2025 

12171 

2318  2464 

2610 

146 

297 

2756 

2903 

3049 

3195 

8341 

3487  '  3638 

3779  8925 

4071 

146 

298 

4216 

4362 

4508 

4653 

4799 

4944  5090 

52:i5  5381 

5526 

146 

299 

5671 

5816 

5962 

6107 

6252 

6397  6542 

6687  6832 

6976 

145 

300 

7121 

7266 

7411 

7555 

7700 

7844  7989 

8138  8278 

8422 

145 

301 

8566 

8711 

8855 

8999 

9143 

9287  9431 

9575  9719 

9863 

144 

302 

480007 

0151 

0294 

0438 

0582 

0725  0869 

1012 

1156 

1299 

144 

303 

1443 

1586 

1729 

1872 

2016 

2159  :  2802 

2445 

2588 

2731 

143 

304 

2874 

3016 

3159 

3302 

3445 

3587  :  3730 

3872 

4015 

4157 

143 

305 

4300 

4442 

4585 

4727 

4869 

5011  5158 

5295 

5487 

5579 

142 

306 

5721 

5863 

6005 

6147 

6289 

6430  1  6572 

6714 

6855 

6997 

142 

307 

7138 

7280 

7421 

7563 

7704 

7845  ;  7986 

8127 

8269 

8410 

141 

308 

8551 

8692 

8833 

8974 

9114 

9255  9396 

9537 

9677 

9818 

141 

309 

9958 

..99 

.239 

•380 

•520 

.661  .801 

.941 

1C81 

1222 

140 

310 

491362 

1502 

1642 

1782 

1922 

2062  2201 

2341 

2481 

2621 

140 

311 

2760 

2900 

3010 

3179 

3319 

3458  '  3597 

3737 

8876 

4015 

lg9 

312 

4155 

4294 

4433 

457214711 

4850 

4989 

5128 

5267 

5406 

139 

313 

5544 

5683 

5822 

5960  i  6099 

6288 

6376 

6515 

6653 

6791 

139 

314 

6930 

7068 

7206 

7344  7488 

7621 

7759 

7897 

8035 

8173 

138 

315 

8311 

8448 

8586 

8724 ' 8882 

8999 

9137 

9275 

9412 

9550 

138 

316 

9687 

9824 

9962 

..99  i  -286 

.374 

.511 

•648 

•785 

•922 

137 

317 

501059 

1196 

1833 

1470  1 1607 

1744 

1880 

2017 

2154 

2291 

137 

318 

2427 

2564 

2700 

2887  2978 

3109 !  8246 

8382 

8518 

3655 

1£6 

319 

3791 

3927 

4033 

4199  i  4385 

4471 

4607 

4743 

4878 

5014 

186 

320 

5150 

5283 

5421 

5557  5698 

5828 

5964 

6099 

6234 

6870 

136 

321 

6505 

6640 

6776 

6911  7046 

7181 

7316 

7451 

7586 

7721 

135 

322 

7856 

7991 

8126 

8260  8395 

8530  i  8664 

8799 

8934 

9068 

135 

323 

9208 

9387 

9471 

9606  9740 

9874!  ...9 

.143 

•277 

.411 

134 

324 

510545 

0679 

0818 

094711081 

1215 

1349 

1482 

1616 

1750 

134 

325 

1883 

20l7 

2151 

2284 ' 2418 

2551 

2684 

2818 

2951 

3084 

133 

326 

3218 

3351 

3484 

3617 1 3750 

3883 

4016 

4149 

4282 

4414 

133 

327 

4548 

4681 

4813 

4946  5079 

5211 

5344 

5478 

5609 

5741 

IS'oi 

828 

5874 

6006 

6139 

6271  6403 

6535 

6668 

6800 

6932 

7064 

U'2 

329 

7196 

7328 

7460 

7592  1  7724 

7855 

7987 

8119 

8251 

8882 

m 

330 

8514 

8646 

8777 

8909 

9040 

9171 

9303 

9434 

9566 

9697 
9 

131 

'sr 

0 

1 

2 

3 

^ 

5 

6 

7 

.^_ 

3j 

LOGARITHMS.  387 

Solution. — Pa2:e  386  is  one  page  of  a  table  of  logarithms  giving 
the  logarithms  of  numbers  from  1  to  10,000  directly^  and  from  which 
the  logarithms  of  other  numbers  can  also  be  found  with  little  trouble. 
Thus,  let  it  be  required  to  find  the  logarithm  of  325.  Now,  as  the 
logarithm  of  100  is  2.  and  of  1000  is  3,  the  logarithm  of  325  must 
be  between  2  and  3,  i  e.  3  and  a  fraction.  The  fractional  part  is 
all  that  is  given  in  the  table,  as  the  integral  can  be  known  bj'  simple 
inspection.  Looking  in  the  table  down  the  column  marked  N 
(numbers),  we  find  325,  and  opposite  it  in  the  column  headed  0,  we 
find  1883,  but  just  above  this  we  observe  51,  which  belongs  to  this 
logarithm  and  which  is  simply  omitted  to  save  space  in  the  table, 
since  it  really  belongs  as  a  prefix  to  all  the  logarithms  clear  down  to 
the  number  332  where  it  is  replaced  by  52.  Prefixing  the  51  to  the 
1883,  we  have  .511883  as  the  decimal  part  of  the  logarithm.  Hence 
log.  325  is  2.511883.  In  like  manner  the  logarithm  of  any  number 
represented  hy  three  Jigures  is  found  from  the  table. 

To  find  the  logarithm  of  a  number  re])resented  by  four  figures.  Let  it 
be  required  to  find  the  logarithm  of  2936.  Looking  for  293  (the 
first  three  figures)  in  the  column  of  numbers,  and  then  passing  to  the 
right  until  reaching  the  column  headed  6,  the  fourth  figure,  we  find 
7756,  to  which  prefixing  the  figures  46,  which  belong  to  all  the 
logarithms  following  them  till  some  others  arc  indicated,  we  have 
for  the  decimal  part  of  the  logarithm  of  2936,  .467756.  But,  as  3  is 
the  logarithm  of  1000,  and  4  of  10,000,  log.  29:56  is  3  and  this  decimal, 
or  log.  2936  :^  3.467756. 

Tofimi  the  logarithm  of  a  number  represented  by  moi-e  than  A^  figures. 
Let  it  be  required  to  find  the  logarithm  of  2845672.  Finding  the 
decimal  part  of  logarithm  of  the  first  4  figures  2845,  as  before,  we 
find  it  to  be  .454082.  Now^  the  logarithm  of  2846  is  153  (millionths, 
really)  more  than  that  of  2845.  Hence,  assuming  that  if  an  increase 
of  the  number  by  1000  makes  an  increase  in  its  logarithm  of  153,  an 
increase  of  672  in  the  number,  will  make  an  increase  in  the  logarithm 
of -jVy2^  or  .672  of  153,  or  103,  omitting  lower  orders,  and  adding 
this  to  .454082,  we  have  .454185  as  the  decimal  part  of  log.  2845672. 
The  integral  part  is  6,  since  2845672  lies  between  ttie  6th  and  7tb 
powers  of  10.     Hence  log.  2845672  =  6.454185.     Q.  e.  d. 

Scholium  I. — If  in  seeking  the  logarithm  of  any  number  any  of 
the  dots  noticed  in  the  table  are  passed,  their  places  are  to  be 
filled  with  O's,  and  the  first  two  figures  of  the  decimal  of  the  loga- 
rithm taken  from  the  0  column  in  the  line  b^low.     Thus  log.  3166  is 


388  LOGARITHMS. 

3.500511.     This  arrangement  of  the  table  is  a  mere  matter  of  con- 
venience to  save  space. 

Scholium  2.— The  column  marked  D  is  called  the  column  of 
Tabular  Difierences ;  and  any  number  in  it  is  the  difference  between 
the  logarithms  found  in  columns  4  and  5,  which  is  usually  the  same 
as  between  any  two  consecutive  logarithms  in  the  same  horizontal 
line.  The  assumption  made  in  using  this  difference ;  viz.,  that  the 
logarithms  increase  in  the  same  ratio  as  the  numbers,  is  only  approxi- 
mately true,  but  still  is  accurate  enough  for  ordinary  use. 

128,  The  Integral  Part  of  a  logarithm  is  called  the 
Characteristic,  and  the  decimal  part  the  Mantissa. 

12i).  Prop. — Tlie  Mantissa  of  a  decimal  fraction, 
or  of  a  mixed  number,  is  the  same  as  the  mantissa 
of  the  number  considered  as  integral. 

Demonstration.— Above  it  was  found  that  log.  2845673=6.454185. 
Now  this  means  that  l08-4  5  4i  8  5  ^  3845673.  Dividing  by  10  suc- 
cessively we  have 


105-454185  — 

284567.3,     or  log. 

284567.2 

=  5.454185, 

10*-454185  — 

38456.73    or  log. 

38456.73 

=  4.454185, 

j^03-454185  — 

3845.673   or  log. 

3845.673 

=-  3.454185. 

102-454t85  _ 

284.5673   or  log. 

284.5673 

--=  3.454185, 

;|0l. 454185  — 

28.45673  or  log. 

38.45673 

=   1.454185, 

100-484185  _ 

3.845673  or  log. 

3.845673 

=  0.454185. 

Now  if  we  continue  the  operation  of  division,  only  writing  0.454185 
—  1,  T.454185,  meaning  by  this  that  the  characteristic  is  negative 
and  the  mantissa  positive,  and  the  subtraction  not  performed,  we 
have 

10T.454185  =  .3845673,      or  log.  .3845672      =  T.454185, 
107.454185  =  .03845673,    or  log.  .03845672    =^.454185, 
lOir.4  5  418  6  —  .003845673,  or  log.  .003845673  =  3.454185, 
etc.      Q.  E.  D. 

Scholium. — The  characteristic  of  an  integral  number,  or  of  a 
mixed  integral  number  and  decimal,  is  one  less  tliau  the  number  of 
integral  places,  as  will  appear  by  comparing  such  number3  with  the 


LOGARITHMS.  889 

powers  of  10,  as  is  done  in  cleinonHtrating  (126).  The  character- 
istic of  a  number  entirely  decinial  fractional,  is  negative,  and  one 
greater  than  the  number  of  O's  immediately  following  the  decimal 
point,  as  appears  from  the  last  demonstration,  or  as  appears  from 
the  fact  that  10->=T'ff  =  -l;  10-«  =  t^  =  .01;  10~^  =  j^^^ 
=  .001 ;  etc. 

EXAMPLES. 
Find  the  logarithms  of  the  following  numhers  :    285  ; 
3145;    29056;    30942;    298.026;    32.56;    2.864;    .3205; 
.00317  ;  .00000328. 
Results,  log.  298.026  =  2.474254  ;  log.  .00317  =3.501059. 


130.  Prob. — To  find  a  number  corresponding  to  a 
given  logarithm. 

Solution. — Let  it  be  required  to  find  the  number  corresponding 
to  the  logarithm  5.515264.  Looking  in  the  table  for  the  next  less 
mantissa,  we  find  .515211,  the  number  corresponding  to  which  is 
8275  (no  account  now  being  taken  as  to  whether  it  is  integral,  frac- 
tional or  mixed ;  as  in  any  case  the  figures  will  be  the  same).  Now, 
from  the  tabular  difference,  in  column  D,  we  find  that  an  increase  of 
133  (millionths,  really)  upon  this  logarithm  (.515211),  would  make 
an  increase  of  1  in  the  number,  making  it  3276.  But  the  given  loga- 
rithm is  only  53  greater  than  this,  hence  it  is  assumed  (though  only 
approximately  correct)  that  the  increase  of  the  number  is  -^Yy  of  1, 
or  53-^133  =  .3984 -h.  This  added  (the  figures  annexed)  to  3275, 
gives  32753984  +  .  The  characteristic,  being  5,  indicates  that  the 
number  lies  between  the  5th  and  6th  powers  of  10,  and  hence  has 
6  integral  places.     .'.  5.515264  =  log.  327539.84  +  . 

EXAMPLES. 

Find  tho  numbers  corresponding  to  the  following  loga- 
rithms :  3.467521  ;  ^.467521  :  0.467521  ;  4.520281 ; 
1.520281;  4.520281;  0.  .520281 ;  1.520281;  2.490160; 
2.490160  ;  and  0.490160. 

Resnlfs,  2.490160  =  log.  309. 1435 -f- ;  ^.490160  =  log. 
,030914+;  0.490160  =  log.  3.091435-i-, 


390  LogaHithMS. 

131.  As  logarithms  are  largely  used  to  facilitate  uiiiner- 
ical  computations,  it  is  important  that  the  student  be  able 
to  take  any  formula  representing  such  operations  and  write 
at  once  the  equivalent  logarithmic  operations. 

EXAMPLES. 
Ex.  1.  If  28.035  :  3.2781  : :  3114.27  :  x,  what  logarithmic 

operations  will  find  x  ? 

Suggestion. — The  logarithm  of  the  product  of  the  means  is  the 
sum  of  their  logarithms;  and  the  logarithm  of  the  quotient  of  this 
product  divided  by  the  first  extreme,  is  the  logarithm  of  said  pro- 
duct minus  the  logarithm  of  the  other  extreme.  .".  log.  .r  =  log. 
3.2781 +log.  3114.27-log.  28.035  =  0.515622  +  3.493356-1.447700 
=  2.561278.  Having  a  table  sufficiently  extended,  the  number 
corresponding  to  this  logarithm  could  be  found,  and  would  be  the 
value  of  a?. 

2.  Find  the  product  of  23  14,  by  5.062,  knowing  that  log. 
23.14  is  1.364363,  log.  5.062  is  0.704322,  and  log.  117.1347 
is  2.068685. 

3.  How  is  287  raised  to  the  5th  power  by  means  of  loga- 
rithms ?    How  is  the  5tli  root  extracted  ? 

4.  Extract  the  5th  root  of  31152784.1  by  means  of  loga- 
rithms, knowing  that  log.  31152784.1  =  7.493497. 

Suggestion.— Log.  ^311527843  =  |  log.  31152784.1  =  1.498699. 
The  number,  therefore,  is  31.52  + . 

6.  What  is  the  cube  root  of  30?  Ans.,  3. 107  4-. 

6.  What  is  the  cube  root  of  .03? 

Suggestion.— Log.  .03  ="§.477121.  Now  to  divide  this  by  8,  we 
have  to  bear  in  mind  that  the  characteristic  alone  is  negative,  i.  e., 
2:477121  =  -2 +  .477121,  or  -1.522879.  This  divided  by  3  gives 
-.507626,  or  0— .507626  =  f.492374.  But  a  more  convenient  method 
of  effecting  this  division  is  to  write  for  the  —2,  —3  +  1,  whence  we 
have  for  2.477121,  -3  +  1.477121,  which  divided  by  3  givesT.492374, 
nearly. 

7.  Divide  3.261453  by  2,  by  4,  by  5. 

Last  quotient,  1.4622906. 


E'C^^mi  L 


DIFFERENTIATION. 

[This  subject  is  inserted  as  the  best  method  of  reaching  the  demon- 
stration of  the  Binomial  Formula  and  the  production  of  the  Logarithmic 
Series.  While  it  is  equally  simple,  to  say  the  least,  with  the  old 
method,  it  is  more  direct,  and  gives  the  student  nothing  but  what  is  of 
fundamental  importance  in  subsequent  mathematical  work.] 

132.  In  certain  classes  of  problems  and  discussions 
the  quantities  involved  are  distinguished  as  Constant  and 
Variable. 

133.  A  Constant  quantity  is  one  which  maintains 
the  same  value  throughout  the  same  discussion,  and  is 
represented  in  the  notation  by  one  of  the  leading  letters 
of  the  alphabet. 

134.  Variable  quantities  are  such  as  may  assume  in 
the  same  discussion  any  value  within  certain  limits  deter- 
mined by  the  nature  of  the  problem,  and  are  represented 
by  the  final  letters  of  the  alphabet. 

III. — If  X  is  the  radius  of  a  circle  and  y  is  its  area,  y  —  r.x^,  as  w.^ 
learn  from  Geometry,  t  being  about  3.1416.     Now  if  «,  the  radium, 

Note.— The  preceding;  part  of  this  volume  ftirnlshes  a  coun*e  in  Algebra  quite  as 
full  as  will  be  found  practicable  or  desirable  in  mo!<t  hi^h  tchool^i  and  academies, 
and  is  an  adequate  preparation  for  college.  This  appendix,  selected  from  Olnet's 
Untvebsitt  Algebra,  i«  inserted  for  such  of  the  above  schools  as  desire  a  ftiller 
course,  and  as  adapting  the  book  to  the  needs  of  many  of  our  colleges  which  do  not 
And  it  exi>edient  to  give  as  much  time  to  this  subject  as  is  required  to  master  the 
University  Algebra.  There  is  nothing  in  the  ordinary  college  course  which 
requires  more  Al^bra  than  is  found  in  this  vohune. 


392  DIFFERENTIATION". 

varies,  y,  the  area,  will  vary ;  but  tt  remains  the  same  for  all  valued 
of  X  and  y.  In  this  case  x  and  y  are  the  variables,  and  tt  is 
a  constant. 

Again,  if  y  is  the  distance  a  body  falls  in  time  a;,  it  is  evident 
that  the  greater  x  is,  the  greater  is  y,  i.  e.,  that  when  x  varies  y  varies. 
We  learn  from  Physics  that  y  =  16j*g!r',  for  comparatively  small 
<listances  above  the  surface  of  the  earth.  In  the  expression  y  = 
\Cy^\x^,  X  and  y  are  the  variables,  and  16 iV  is  a  constant.    .-.  y  ooa;'. 

Once  more,  suppose  we  have  y^  —  253:^— Saj'^— 5,  as  an  expressed 
relation  between  x  and  y,  and  that  this  is  the  only  relation  which  is 
required  to  exist  between  them  ;  it  is  evident  that  we  may  give 
values  to  x  at  pleasure,  and  thus  obtain  corresponding  values  for  y. 
Thus  if  2J  =  1,  ?/  =  +  v^TT,  \tx  —  2,y=  ±  ^/ISS,  etc.,  etc.  In  such  a 
case  X  and  y  are  called  variables.  But  we  notice  that  if  we  give  to  x 
such  a  value  as  to  make  3a;'^  +  5>25a^  (as,  for  example,  1,  |,  etc.),  2/ 
will  be  imaginary.  This  is  the  kind  of  limitation  referred  to  in  our 
definition  of  variables. 

13ij»  ScH. — The  pupil  needs  to  guard  against  the  notion  that  the 
terms  constant  and  variable  are  synonyms  for  knovm  and  unknown, 
and  the  more  so  as  the  notation  might  lead  him  into  this  error.  The 
quantities  he  has  been  accustomed  to  consider  in  Arithmetic  and 
Elementary  Algebra  have  all  been  constant.  The  distinction  here 
made  is  a  new  one  to  him,  and  pertains  to  a  new  class  of  problems 
and  discussions. 

130.  A  Function  is  a  quantity,  or  a  mathematical 
expression,  conceived  as  depending  for  its  value  upon  some 
other  quantity  or  quantities. 

iLTi. — A  man's  wages  for  a  given  time  is  a  function  of  the  amount 
received  per  day,  or,  in  general,  his  wages  is  a  function  of  the  time 
he  works  and  the  amount  he  receives  per  day.  In  the  expression 
/  =  IGjJjic'  (i.V4),  second  illustration,  y  is  a  function  of  x,  i.e.,  the 
ipace  fallen  through  in  a  function  of  the  time.  The  expression 
2ax^—Sx-\-  5b,  or  any  expression  containing  x,  may  be  spoken  of  as  a 
function  of  x. 

137.  When  wo  wish  to  indicate  that  one  variable,  as  y, 
is  a  function  of  another,  as  x.  and  do  not  care  to  be  more 
specific,  we  write  y  =  f{x),  and  read  "^Z  equals  (or  is)  a 


DIFFRRKNTIATION.  393 

function  of  a:."'  This  means  nothing  more  than  that  y  is  equal 
to  some  expression  containing  the  varia))le  r,  and  which 
besides  may  contain  any  constants.  If  we  wish  to  indicate 
several  different  expressions  each  of  which  contains  x,  wc 
write/(.r),  q)  (2:),  or/  (;r),  etc,  and  read  '*  the/ function  of 
x"  "  the  q>  function  of  re,"  or  '*  the  /'  function  of  ic." 

III. — The  expression /(;r)  may  stand  for  0^—2^  +  5,  or  for  3(/z'— rr^V 
or  for  any  expression  containing  r  wmbined  in  any  way  with  itself  01 
with  constants.  But  in  the  same  discussion  f{x)  will  mean  the  same 
thing  tlirougliout.  So  again,  if  in  a  particular  discussion  we  have  a 
certain  expression  containing  X{e.  g.,  %x^—ax-\-Zab).  it  may  be  repre- 
sented by  /(ir),  while  some  other  function  of  x,  c.  g.,5{n*—'j*)  +  2x'^ . 
might  be  represented  by/  (x),  or  0  (x). 

138.  In  equations  expressing  the  relation  betweeen  two 
variables,  as  in  if  =  daafi—x^,  it  is  customary  to  speak  of 
one  of  the  variables,  as  y,  as  a  function  of  the  other,  x. 
Moreover,  it  is  convenient  to  think  of  x  as  varying  and 
thus  producing  change  in  //.  When  so  considered,  x  is 
called  the  Independent  and  y  the  Dependent  varial)le.  Or 
we  may  speak  of  y  as  a  function  of  the  variable  x. 

130.  An  Inftnitesininl  is  a  quantity  conceived 
under  such  a  form,  or  law,  as  to  be  necessarily  less  than 
any  assignable  quantity. 

Infinitesimals  are  the  increments  by  which  continuous 
number,  or  quantity  (H),  may  be  conceived  to  change  value, 
or  grow. 

III.— TYwe  affords  a  good  illustration  of  continuous  quantity,  or 
number.  Thus  a  period  of  time,  as  5  hours,  increases,  or  grows,  to 
another  period,  as  7  hours,  by  intinitesimal  increments,  i.  t.,  not  by 
hours,  minutes,  or  even  seconds,  but  by  elements  which  are  less  than 
any  assignable  quantity. 

140.  Consecutive  Values  of  a  variable  are  values 
which  differ  from  each  other  by  less  than  any  assignable 
quantity,  i.  e.,  by  an  infinitesimal.     Consecutive  values  of  a 


394  DIPFERENTIATIOK. 

function  are  values  which  correspond  to  consecutiye  values 
of  its  variable. 

141,  A  Differential  of  a  function,  or  variable,  is  the 
difference  between  two  consecutive  states  of  the  function, 
or  variable.     It  is  the  same  as  an  infinitesimal. 

Ii^L- — Resuming  the  illustration  y  =  IQ^x^  (/'Vt^),  let  x  be  thought 
of  as  some  particular  period  of  time  (as  5  seconds),  and  y  as  the  dis- 
tance through  which  the  body  falls  in  that  time.  Also,  let  x'  represent 
a  period  of  time  only  infinitesimally  greater  than  x,  and  y'  the  distance 
through  which  the  body  falls  in  time  x' .  Then  x  and  x'  are  consecu- 
tive values  of  x,  and  y  and  y'  are  consecutive  values  of  y.  Again,  the 
difference  between  x  and  x',  as  cc'— .r,  is  a  differential  of  the  variable 
X,  and  y'—y  is  a  differential  of  the  function  y. 

142,  Notation. — A  differential  of  x  is  represented  by 
writing  the  letter  d  before  x,  thus  dx.  Also,  dy  means,  and 
is  read  "differential  y.^' 

Caution. — Do  not  read  dx  by  naming  the  letters  as  you  do  ax ;  but 
read  it  "  differential  x."  The  d  is  not  a  factor,  but  an  abbreviation  for 
the  word  differential. 

143,  To  Differentiate  a  function  is  to  find  an 
expression  for  the  increment  of  the  function  due  to  an 
infinitesimal  increment  of  the  variable;  or  it  is  the  process 
of  finding  the  relation  between  the  infinitesimal  increment 
of  the  variable  and  the  corresponding  increment  of  the 
function. 

RULES    FOR    DIFFERENTIATING. 

144,  Rule  I. — To  dijferentiate  a  single  variable, 
sim'ply  write  the  letter  d  before  it. 

This  is  merely  doing  what  the  notation  requires.  Thus,  if  x  and 
x'  are  consecutive  states  of  the  variable  x,  ^.  e.,  if  x  is  what  x  becomes 
when  it  has  taken  an  infinitesimal  increment,  x'—x  is  the  differential 
of  X,  and  is  to  be  written  dx.  In  like  manner,  y'  —y  is  to  be  written 
dy,  y'  and  y  being  consecutive  values. 


DIFFERENTIATION.  395 

145.  Rule  II. — Constant  factors  or  divisors  ap- 
pear in  the  differential  tJie  same  as  in  the  function. 

Dem. — Let  ua  take  the  function  y  =  ax,  in  which  a  is  any  constant, 
integral  or  fractional.  Let  x  take  an  infinitesimal  increment  dx, 
becoming  x  +  dx  \  and  let  dy  be  the  corresjxjnding  "  increment  of  y,  so 
that  when  x  becomes  x-k-dx,  y  becomes  y  +  dy.     We  then  have 

1st  state  of  the  function     .    .  y  —  ax\ 

2d,  or  consecutive  state     .    .    y  +  dy  :=  a(x+dx)  =  ctx  +  adx. 

Subtracting  the  Ist  from  the  2d  dy  =  adx, 

which  result  being  the  difference  b3tween  two  consecutive  states  of 

the  function,  is  its  differential  {141).     Now  a  appears  in  the  difTei- 

ential  just  as  it  was  in  the  function.     This  would  evidently  be  the 

1 
same  if  a  were  a  fraction,  as  -  .    We  should  then  have,  in  like  man- 
m 

ner,  dy  =  —  dx,  for  the  differential  of  y  =  —  a;.    Q.  E.  D. 


146,  Rule  III. — Constant  terms  disappear  in  dif- 
ferentiating ;  or  the  differential  of  a  constant  is  0. 

Dem. — Let  us  take  the  function  y  =  ax-\-b,m  which  a  and  &  are 
constant.  Let  x  take  an  infinitesimal  increment  and  become  x  +  dx  ; 
and  let  dy  be  the  increment  which  y  takes  in  consequence  of  this 
change  in  x,  so  that  when  x  becomes  x  +  dx,  y  becomes  y  +  dy.  We 
then  have 

1st  state  of  the  function     .     .  y  =  ax  +  b; 

2d,  or  consecutive  state     .    .    y+dy  =  a(x+dx)  +  b  =  ax+ada+b. 

Subtracting  the  1st  from  th  •  2d  dy  =  adx, 

which  being  the  difference  between  two  consecutive  states  of  the 
function,  is  its  differential  {141).  Now  from  this  differential  the 
constant  b  has  disappeared. 

We  may  also  say  that  as  a  constant  retains  the  same  value,  there 
is  no  difference  between  its  consecutive  states  (properly  it  has  no  con- 
secutive states).  Hence  the  differential  of  a  constant  may  be  spoken 
of  as  0.     Q.  B.  D. 


*  The  word  "  contemporaneoaB  "  is  often  used  in  Ibis  connection. 


o9v 


DIFPERENTIATIGK. 


147,  Rule  IV.  -—  To  differentiate  the  algebraic 
sum  of  several  variables,  differentiate  each  term  sep^ 
arately  and  connect  the  differentials  with  the  same 
signs  as  the  terms. 

Dem. — Let  u  =  x+y—z,  u  representing  the  algebraic  sum  of  the 
variables  x,  y,  and  —z.  Then  is  du  —  dx-^dy—dz.  For  let  dx,  dy, 
ind  dz  be  infinitesimal  increments  of  x,  y,  and  z  ;  and  let  du  be  the 
increment  which  u  takes  in  consequence  of  the  infinitesimal  changes 
in  X,  y,  and  z.    We  then  have 

1st  state  of  the  function ....  u  =  x-^y—z  ; 

2d,  or  consecutive  state  .     .     .    .    u  +  dii  z=  x-^dx-\-y^dy—{z-\-dz). 

^f u  +  du  =  x  +  dx  +  y  +  dy—z—dz. 

Subtracting  the  1st  state  from  the  3d      du  =  dx+dy—dz.    q.  e.  d. 


148.  Rule  v.  — The  differential  of  the  product 
of  two  variables  is  the  differential  of  the  first  into 
the  second,  pias  the  differential  of  the  second  into  the 
first. 

Dem. — Let  i*  =  jcy  be  the  first  state  of  the  function.  The  consecu- 
tive state  is  w  +  dii  =  {x  +  dx)  {y + dy)  —  xy+ ydx  +  xdy  +  dx  •  dy.  Sub- 
trading  the  1st  state  from  the  consecutive  state  we  have  the  differ- 
ential, i.  e.,du  =  ydx  +  xdy  +  dx  •  dy.  But,  as  dx  •  dy  is  the  product  of 
two  infinitesimals,  it  is  infinitely  less  than  the  other  terms  (ydx  and 
xdy),  and  hence,  having  no  value  as  compared  with  them,  is  to  be 
dropped.*    Therefore,  du  =  ydx  +  xdy.    Q.  e  D. 


*  It  will  doubtless  appear  to  the  pupil,  at  first,  as  if  this  {?;ave  a  result  only 
approximately  correct.  Such  is  not  the  fact.  The  result  is  absolutely  correct.  J\^o 
error  is  introduced  by  dropping  dx  •  dy.  In  fact  this  term  mnst  be  dropped  accord- 
ins  to  the  nature  of  infinitesimals.  Notice  that  by  definition  a  quantity  which  is 
Infinitesimal  with  respect  to  another  is  one  which  has  no  assignable  magnitude 
with  reference  to  that  other.  Hence  we  must  so  treat  it  in  our  reasoning.  Now 
dx-dy  is  an  infinitesimal  of  an  infinitesimal  (i.e.,  two  infinitesimals  multiplied 
together),  and  hence  is  infinitesimal  with  reference  to  ydx  and  xdy,  and  must  b*» 
treated  as  having  no  assignable  value  with  respect  to  them;  that  is,  it  must  b« 
dropped. 


DIFFERENTIATION.  397 

149.  Rule  VI. — The  differential  of  the  product 
of  several  variables  is  the  sum  of  the  products  of  the 
differential  of  each  into  the  product  of  all  the 
others. 

Dem.— Let  u  —  xyz\  then  du  =  pzdx  +  xzdy  +  xydz.  For  the  Ist 
state  of  the  function  is  u  —  xyz,  and  the  2d,  or  conaecutive  state. 
H^da  =  (a?  +  dx)  {y  +  dy)  [z  +  dz),  or  n  +  du  —  xyz  +  yzdx  +  xzOy  +  xydz  + 
xdydz  +  ydxdz+zdxdy  +  dxdydz.  Subtracting,  and  dropping  all  infini- 
tesimals of  infinitesimals  (see  preceding  rule  and  foot-note),  we  have 
du  =  yzdx  -f  xzdy  +  xydz. 

In  a  similar  manner  the  rule  can  be  demonstrated  for  any  number 
of  variables.     Q.  E.  D. 


150.  Rule  VU.—TJie  differential  of  a  fraction 
having  a  variable  numerator  and  denominator  is  tlie 
differential  of  the  numerator  multiplied  by  the 
denominator,  minus  the  differential  of  the  denom- 
inator  multiplied  by  the  numerator,  divided'  by  the 
square  of  the  denominator. 

Dem. — Let  M  =  - :  then  is  du  = ~  .     For,  clearing  of  f rao- 

y  y' 

tions,  yu  =  x.     Differentiating  this  by  Rule  5th,  we  have  udy  +  ydu  — 

X  xdv 

dx.      Substituting  for  u  its  value  -,  this  becomes  — ^  +  ydu  =  dx. 

y  y 

ydx — xdv 
Finding  the  value  of  du,  we  have  du  = f-^.    <i.  E.  d. 

lol.  Cor. — TJie  differential  of  a  fraction  havirig  a  con- 
slant  numerator  and  a  variable  denominator  is  the  product 
of  the  numerator  with  its  sign  changed  into  the  differeiitial 
of  the  denominator^  divided  by  the  square  of  the  denom- 
inator. 

Let  u  =  -.    Differentiating  this  by  the  rule  and  calling  the  dif- 

y 


398  DIEFERENTIATIOK. 

ferential  of   the  constant  {a)  0,  we   have  du  =  ^JZL^^  =  -ady 

Q.  E.  D. 

Iii2,   SCH. — If  the  numerator  is  variable  and  the   denominator 
constant,  it  falls  under  Rule  3. 


153  •  Rule  VIII. — The  differential  of  a  variable 
affected  with  an  exponent  is  the  continued  product 
of  the  exponent,  the  variable  ivith  its  exponent 
diminished  by  1,  and  the  differential  of  the  variable. 

Dem. — 1st.  When  the  exponent  is  a  positive  integer.  Let  y  =  x"% 
m  being  a  positive  integer ;  then  dy=m.x"'-^dx.  For  y=x'^=x  .x.x.x. 
to  m  factors.  Now,  differentiating  this  by  Rule  6,  we  have  dy  = 
(xxx  .  .  to  m—1  factors)  dx  +  {xxx  .  .  to  m—1  iactors)dx+  etc.,  to  m 
terms;  or  dy  =  x"'-^dx  +  x"'-^dx  +  x'''-^dx+  etc.,  to  m  terms.  There- 
fore dy  =  map-^dx. 

m 

2d.  When  the  exponent  is  a  positive  fraction.     Let  y  =  x"  ,  "^  being 

a  positive  fraction  ;  then  dy  =  ^a?*     dx.    For  involving  both  members 
to  the  nth.  power  we  have  y"  =  x'".    Differentiating  this  as  just  shown, 

we  have   ny"-^dy  =  mx'"-^dx.     Now   from  y  =  x^'  we  have  y"-'  = 

win — m  ,nn — m 

X    "    .     Substituting  this  in  the  last  it  becomes  nx    '*    dy=mQf^^dx ; 

•iiifi—m  m 

whence  dy  —  ^cc  "   dx  =  ^a5»     dx.    q.  e.  d. 

3d.  Wh^n  the  exponent  is  n£gative.     Let  y  =  x-",   n  being  integral 
or  fractional ;  then  dy  =  —n'X-"~^dx.    For  y  =  ar-"  =  — ,  which  dif- 

ferentiated  by   Rule  7,  Cor.,  gives  dy  z= —  =  —  nx-^-^dx. 

Q.  B.  D. 


EXAMPLES. 

1.  Differentiate  y  =  3a;2— 22^  +  4. 

Solution,— The  result  is  dy  —  Qxdx  —  ^dx.     Which  is  thus 


DIFFERENTIATION.  399 

obtained  :  By  Rule  1,  the  dilierential  of  y  is  dy.  To  differentiate  the 
second  member  we  differentiate  each  term  separately  according  to 
Rule  4  In  differentiating  3x',  we  observe  that  the  factor  3  is 
retained  in  the  differential,  Rule  3,  and  the  differential  of  x'  is,  by 
Rule  8,  %xda.  Hence,  the  differential  of  %j?  is  6iwir.  The  differential 
of  —2a*  is  —%dx.  By  Rule  3,  the  constant  4  disappears  from  the  dif 
ferential,  or  its  differential  is  0. 

2.  Differeutiate  y  ==  2ax^-{-4:a3^—x-\-7n. 

Res2iU,  dy  =  4:axdx-^12axMx—dx. 

3.  Differentiate  y  =  ob3^—'d0x^-^4:X. 

4.  Differentiate  y  =::  Aa^  +  B^^  a-  Cx*. 

134:,  ScH. — It  is  desirable  that  the  pupil  not  only  become  expert 

in  writing  out  the  differentials  of  such  expressions  as  the  above,  but 
that  he  know  what  the  operation  signifies.  Thus,  suppose  we  have 
the  equation  y  =  5j.  This  expresses  a  relation  between  x  and  y. 
Now,  if  X  changes  value,  y  must  change  also  in  order  to  keep  the 
equation  true.  In  this  simple  case  it  is  easy  to  see  that  y  must 
change  5  times  as  fast  as  x  in  order  to  keep  the  equation  true. 
This  is  what  differentiation  shows.  Thus,  differentiating,  we  have 
dy  =  Mx.  That  is,  if  x  takes  an  infinitesimal  increment,  y  takes  an 
infinitesimal  increment  equal  to  5  times  that  which  x  takes  ;  or,  in 
other  words,  y  increases  5  times  as  fast  as  x. 

Now  let  us  take  a  case  which  is  not  so  simple.  Let  y=h£^—)lx-^A, 
and  let  it  be  required  to  find  the  relative  rate  of  change  of  x  and  y. 
Differentiating,  we  have  dy  —  Qxdx—'Zdx  —  (6a;— 2)  dx.  This  shows 
that,  if  X  takes  an  infinitesimal  increment  represented  by  dx,  y  takes 
one  (represented  by  dy)  which  is  Ga;— 2  times  as  large  ;  i.  «.,  that  y 
increases  6a;— 2  times  as  fast  as  x.  Notice  that  in  this  case  the  rela 
tive  rate  of  increase  of  x  and  of  y  depends  on  the  value  of  x.  Thus, 
when  a;  =  1,  y  is  increasing  4  times  as  fast  as  x  ;  when  a;  =  2,  y  is 
increasing  10  times  as  fast  as  x  ;  when  a;  =  3,  y  is  increasing  16  times 
as  fast  as  x  ;  etc. 

5.  Differentiate  y  =  3^—3^,  and  explain  the  significance 
of  the  result  as  above.  Resulty  dy  =  (5a:*— 3.c2)  dx. 

6.  In  order  to  keep  the  relation  2y  =  ^x^  true  as  x  varies, 
how  must  y  vary  in  relation  too;?     What  is  the  relative 


400  DIFFERENTIATIOM". 

rate  of  change  when  a:  =  4  ?    When  .r  ~  2?    When  a:  =  1  ? 

When  a;  =  i?     When./.=:i? 

Answers.  When  x  =  4:,  y  increases  iJi  times  as  fast  as  x. 

When  X  =z  ^,  y  increases  at  the  same  rate  as  x.    In  general 

y  increases  3x  times  as  fast  as  x.     When  x  is  less  than  -J,  y 

m  creases  slower  than  x. 

2x^  x^ 1 

7  to  12.  Differentiate  the  following :  «^  =  ^r— ;  w  =  -^ — r  : 

°  Zy  x^-\-\ 

y  —  xh^;    u  =  a^y^-^-Qx;    y  —  ar"^— 8ar*4-4a;8— a^l ;    and 
y:z^\:^-\7?^x. 

13  to  17.    Differentiate    y  —  {d^-^-x?)^',     y  =i  {a-\-'jp)^ -, 

y  =  (3a;-2)*;  y  =  (2-a;2)-2;  and  y  =  (l+x)"^. 

Sug's. — Such  examples  should  be  solved  by  considering  the  entire 
quantity  within  the  parenthesis  as  the  variable.  This  is  evidently 
admissible,  since  any  expression  which  contains  a  variable  is  variable 

when  taken  as  a  whole.    Thus  to  differentiate  y  =  (a  +  x^)'^,  we  take 
the  continued  product  of  the  exponent  (I),  the  variable  (a  +  a?-)  with 

its  exponent  diminished  by  1,  [i.  e.,  {a  +  x^r^],  and  the  differential  of 
the  variable  (i.  e.  the  differential  of  «4-a;*,  which  is  2ixdx\    This  gives 

us  dy  =  \{a-\-  x^)~^2xdx,  or  dp  —  ^x(a  +  x^)  ^dx  = — —  . 

8/^a+ar* 

1-1  1 

18  to  22.    Differentiate 


1+a:'     (l+a^)^'     (l+a;)^' 


1  A  1 

—m-i r?;    and    —m 


(1  +a:)«'  (l  +  x)^ 

23.  In  the  expression  6x^,  when  x  is  greater  than  1  does 
the  function  (6x^)  change  faster  or  slower  than  x?     How, 
when  X  is  less  than  |?     What  does  the  process  of  differ 
entiating  6a;3  signify? 

Answer  to  the  last.  Finding  the  relative  rate  of  change  of  6a^ 
and  of  X,  or  finding  what  increment  Ci*  takes  when  x  takes  the 
increment  dx. 


INDETERMINATE    C0EFFICIKNT8.  401 

Or,  in  fitill  otber  words,  finding  tlie  diflerence  between  two  con- 
secutive states  of  C}X^,  and  hence  the  relation  betwetn  an  infinitesimal 
increment  of  x  and  the  corresponding  increment  of  6x*. 


INDETERMINATE     COEFFICIENTS. 

lo5»  Indetefininate  Coefficients  are  coefficients 
ftssumed  in  the  demonstration  of  a  theorem  or  the  solution 
of  a  problem,  whose  values  are  not  known  at  the  outset,  but 
are  to  be  determined  by  subsequent  processes. 


ISO.  Prop.— If  A  +  Bx  +  Cx2  4-  Dx^  +  etc.  =  A'-f 
B'x+C'x^-|-D'x3-f-  etCf  in  which  x  is  a  variable  ajid  the 
coefficients  A,  B,  A'  B',  etc.,  are  constants,  tlie  coefficients  of 
the  like  powers  of  x  are  equal  to  each  other.  That  is,  A  = 
A'  (these  being  the  coefficients  of  :sP),  B  =  B',  C  =  0',  etc. 

Dem. — Since  the  equation  is  true  for  any  value  of  x,  it  is  true  for 
a;  =r  0.  Substituting  this  value,  we  have  A  =  A'.  Now  as  ^  =  ^' 
are  constant,  they  have  the  same  values  whatever  the  value  assigned 
to  X.  Hence  for  any  value  of  a;,  ^  =  A  .  Again,  dropping  A  and  A', 
we  have  Bx  +  Cx'^  +  D^  +  etc.  =  B'x  +  G't?  +  D  x^  +  etc. ,  which  is  true 
for  any  value  of  x.  Dividing  by  x,  we  obtain  B+Cx  +  Dx?+  etc.  = 
B'  +  Gx  +  B'x^  +  etc.,  likewise  true  for  any  value  of  x.  Making  x  =  0, 
B  =  B',  as  before.  In  this  manner  we  may  proceed,  and  show  thaf. 
G=:C,  J)  =  D',  etc.     Q.  E.  D. 

lo7.  Cor.— If  A-f  Bx-f-Cx^-f  Dx3+  etc.  =  0,  is  true 
for  all  values  of  x,  each  of  the  coefficients  A,  B,  C,  etc.,  is  0. 

For  we  may  write  A -{■  Bx ■{- Gx^  +  Dx^  +  Ex*  +  Fx^  +  etc.  =0-\-Ox-i 
Oa;'  -4-  02;«  +  Ox*  +  Ox*  +  etc.  Whence  by  the  proposition  ^  =  0,  B  =  0. 
0  =  0,  etc. 


402       DEMOI^-STBATIOi^   OF  THE  BIl^^OMIAL  FORMULA. 


jCechon  il 


DEMONSTRATION    OF  THE    BINOMIAL  FORMULA. 

liiS.    Theorem, — Whatever  the  value  of  m,  i.   e.. 
tvhether  integral  or  fractional,  positive  or  negative. 

(a+a;)»»  =  a"" -\- ma'^^x -{■  ^v^— J  am^i^^ 

^m(m-l)(m-2)^^^ 

Dem.  — Assume 

(a  +  a;)«  =  ^  +  J5aj+Cic*  +  i>a?*  +  .Ete*+i7!r»-i- etc.  (1) 

in  which  A,  B,  C,  etc.,  are  indeterminate  coeflBcients  independent  of 
X  (t.  e.  constants),  and  are  to  be  determined.  To  determine  these  co- 
efficients we  proceed  as  follows : 

Differentiating  (1),  we  have 

m{a  +  x)'^-^dx  =  Bdx + 2Gxdx  +  SDa^dx + iJSufidx + Fxi^dx  +  etc. 
Dividing  by  dx,  we  have 

m(a  +  cc)"^»  =  B  +  2Cx  +  SDx'-  +  4.5^  +  QFx!*  +  etc.  (3) 

Differentiating  (3)  and  dividing  by  dx,  we  have 

m(m—l)(a  +  x)"-^  =  2(7+  3  •  SJDx  4-  3  •  4^a;2  +  4  •  5Fx!'  +  etc.     (3) 

Differentiating  (3)  and  dividing  1  y  dx,  we  have 
m{m-l)im-2){a  +  x)'^^  =  2  '  SI)  +  2  '  d    4:Ex  +  d    i  -  5Fx^+  etc.   (4) 

Differentiating  (4)  and  dividing  by  dx,  we  have 
m{m-l)(m—2){m-'d)(a  +  ai)^'*-*  =  3  •  8  •  4^+3  •  3  '  4  '  5Fx+  etc.   (i5) 


DEMONSTRATION   OF   THE   BINOMIAL   THEOREM.        403 
Differentiating  (5)  and  dividing  by  dXy  we  have 

w(7n-l)(m-2)(m-3Xm-4Xa  +  a;)— »  =  2  •  8  •  4  •  aF-^  etc.        (6) 

We  have  now  gone  far  enough  to  enable  us  to  determine  the  co- 
eflBcients  A,  B,  C,  D,  E,  and  F,  and  doubtless  to  determine  the  law  of 
the  series. 

As  all  the  above  equations  are  to  be  true  for  all  values  of  x,  and  as 

the  coeflBcients  A,  B,  C,  etc.,  are  constants,  i.  e.,  have  the  same  values 

for  one  value  of  a;  as  for  another,  if  we  can  determine  their  values  for 

one  value  of  x,  these  will  be  their  values  in  all  cases.    Now,  making 

x  =  0,  we  have  from   (1)   A  =1  ,  from  (3),  B  =  m;  from  (3),  C  = 

m(m—l)     -         ,.^    _.      m{m-l)(m-2)     ^         ,^,     „ 
r^ —  ;  from  (4),  D  =  -^ .^ ^ ;    from  (5),   E  = 

m(m-l)(m-2Xm-3) ,  ^ ,^^  w(m  -  IXw  -  2)(77>  -  3)(m  -  4) 

j^ ,  from  (0),  Jf  = — 

These  values  substituted  in  (1)  give 

+  fn(m-l){m-2)im-S)(m-i)  ^.^^  ^ 

Which  18  the  Binomial  Formula. 

ISO,  CoR.  1. — The  nth,  or  general  term  of  the  series  is 
m{m^l){ni-2) {m-n-^2)  ,       , 

For  we  observe  that  the  last  factor  in  the  numerator  of  the  oo- 
eflBcient  of  any  particular  term  is  m—  the  number  of  tbe  term  less 
2,  i.  e.,  for  the    nth  term,    m—(n—2\,  or   m—n  +  2;  and  the  last 


404  THE    LOGARITHMIC    SERIES. 

factor  in  the  denominator  is  the  number  of  the  term  —1,  i,  e.,  for  the 
nth  term,  ji—1.  The  exponent  of  x  in  any  particular  term  is  m— 
the  number  of  the  term  less  1,  i.  e,,  for  the  nth  term,  m—in—X),  or 
m— 7i-h  1 ;  and  the  exponent  of  y  in  any  term  is  one  less  than  the 
number  of  the  terra,  i.  e.,  for  the  7ith  term,  n—l. 

100,  Def. — In  a  series  the  Scale  of  Itelation  is  the 

relation  which  exists  between  any  term  or  set  of  terms  and 
the  next  term  or  set  of  terms. 

101,  Cor.  2. — The  scale  of  relation  in  the  binomial 

series  is  ( — — l)-,  since  the  nth  term  multiplied  hy 

this  produces  the  (n  -f-  \)th  term. 

This  is  readily  seen  by  inspecting  the  series,  or  by  writing:  the 
(n4-l)th  term  and  dividing  it  by  the  wth.  Thus,  substituting  in  the 
general  term  as  given  above,  71  + 1  for  n,  we  have 

tn(m-l)(7n-2) (ra-7i-\-\)^^^_^^^ 

\n 

asthe(7i  +  l)th  term.     This  divided  by  the  nth,  or  preceding  term* 

m — n  +  lt/        /m  +  1      \y 
gives  -,  or  I 1  -  • 


M3TIOM  IIL 


THE     LOGARITHMIC     SERIES. 

10^,  The  Modulus  of  a  system  of  logarithms  is  a 
constant  factor  which  depends  upon  the  base  of  the  system 
and  characterizes  the  system. 

103.  JProp.—The  differential  of  the  logarithm  of  a 


THE    LOOAHITHMIO    '^KRIK!^.  40?) 

number  is  the  differential  of  the  number  multiplied  bif  the 
modulus  of  the  system^  divided  by  the  number  ; 

Ory  in,  the  Napierian  system,  the  modulus  being  1,  the 
differential  of  the  logarithm  of  a  number  is  the  differetitial 
of  the  number  divided  by  the  number. 

Dem.— Let  X  represent  any  number,  t.  e.  be  a  variable,  and  n  be  a 
:;onstant  such  that  y  =  aJ».  Then  log  y  =  »  log  x  (1^3).  Difl'eren 
tiating  p  -  x^,  we  have  dy  =  nx^-^dx ;  whence 

dy 

n  =  -^-~  =  -^-  =1^  =  JL 

a;»-'rf.r      j?»  ,        V  ■,       dx'  (*) 

—  dx      -dx      — 

X  X  X 

Again,  whatever  the  differentials  of  log  y  and  log  x  are,  n  being  a 
constant  factor  we  shall  have  the  differential  of  log  y  equal  to  n  times 
the  differential  of  log  x,  which  may  be  written 

d{\og y)  =  n'd (log  x),  whence  n  =  j^~^        (2.) 

Now  equating  the  values  of  ti  as  represented  in  (1)  and  (2),  we  have 

dy 

d{\og  y)        y  dy 

TTT-  -^  =  rr  ■  Whence  d(logy)  bears  the  same  ratio  to  — ,  as 
d(log;r)       dx  \    &&'  y 

X 

rf(log  x)  does  to  —  .    Let  m  be  this  ratio.    Then  rf(log  y)  = ,  and 

- ,,       .      mdx 
d{logx)  =  -^. 

We  are  now  to  show  that  m  is  constant  and  depends  on  the  base  of 
the  system. 

To  do  this,  take  y  —  v^' ,  from  which  we  can  find  as  above  n'  = 

d^ 

(ogj^  =  ^ .     Now  as  m  is  the  ratio  of  d  (log  y)  to  -^ ,  it  is  also  the 
d  Jog  z)       di  ^    ^ "        y  ' 

9 

ratio  of  d([og  z)  to  -  ;  and  (f  (log  z)  =  — -  .    Thus  we  see  that  in  any 

0  z 

case  in  the  same  system  the  same  ratio  exists  between  the  differen- 
tial of  the  logarithm  of  a  number  and  the  differential  of  the  number 
divided  by  the  number.    Therefore  w  is  a  constant  factor. 


406  THE    LOGARITHMIC    SERIES. 


Affain  -^  —  m-  indicates  the  relative  rate  of  chau<^e  of  a  lo^a- 
^        ax  X 

rithm  and  ita  number.     Now  it  is  evident  that  the  larger  the  base  the 

slower  the  logarithm  will  change  with  reference  to  the  number.    (See 

examples  under  Art  117.)    But  the  factor  -  varies  inversely  in  the 

number  ;  hence  m  must  vary  with,  or  be  a  function  of  the, base.* 


164,  Prob.— To  produce  the  logarithmic  series. 

Solution. — The  logarithmic  series,  which  is  the  foundation  of  the 
usual  method  of  computing  logarithms,  and  of  much  of  the  theory  of 
logarithms,  is  the  development  of  log  (1  +ic).  To  develop  log  (1  +2!), 
assume 

log  {l  +  x)  =  A  +  Bx+Cx^  +  D^-¥E^-¥F3^+  etc.,  (1) 

in  which  a;  is  a  variable,  and  A,  B,  C,  etc.,  are  constants. 
Differentiating  (1),  we  have 
mdx\ 


1  +  x 
Dividing  by  dx. 


Bdx  +  2Gxdx+dI)x^dx  +  ^m^dx+5me*dx+  etc 


^         B  +  2Cx+SDx'  +  4^af^  +  5If\cU  etc.  (8) 


1  +x 
Differentiating  (3),  and  dividing  by  dx,  we  have 

-m^P-i— =2(7+2. 32)a;+ 3. 4^2  +  4. 5i?'a.'3+  etc  (8) 

Differentiating  (3),  and  dividing  by  2  and  by  dx,  we  have 

W-— i--  =  3i)  +  3.4^a;+2-3-5i^2^.  etc  (4) 

(1  +xf  ^  ' 

Differentiating  (4),  and  dividing  by  3  and  dx,  we  have 

-7»  — L_  =  4^+4  5Fx+  etc  (5) 

(1  +  xf  ^  ^ 

*  What  the  relation  of  the  modulus  to  the  base  is,  we  are  not  now  concemed  to 
know ;  it  will  be  determined  hereafter. 

t  The  number  is  1+aj;  hence  the  differential  is  m  times  the  differential  of  1-f » 
divided  by  the  number  1+x. 


THE    LOGARITnMTC    SERIES.  407 

Differentiating  (5),  and  dividing  by  4  and  dx,  we  have 

m  TT-i-i  =  5F+  etc.*  (6) 

(I  +  «)» 

We  have  now  gone  far  enough  to  enable  us  to  determine  the  coef- 
ficients A,  B,  G,  I),  E,  and  F,  and  these  will  probably  reveal  the  law 
of  the  series. 

As  all  the  above  e<iuationB  are  to  be  true  for  all  values  of  a*,  and  as 
the  coeflBcients  .1,  B,  C,  etc.,  are  constant,  i.  e.,  have  the  same  values 
for  one  value  of  x  as  for  another,  if  we  can  determine  their  values  for 
one  value  of  x,  these  will  be  their  values  in  all  cases.  Now,  making 
a;  =  0,  we  have,  from  (1),  A  =  log  1  =  0;  from  (2),  B  =  tn  ;  from  (3), 
G=—lm;  from  (4),  D  =  lm;  from  (5),  E=  —\m\  from  (6),  F=\m. 
These  values  substituted  in  (1)  give 

y^       ^       ^       y* 

log  ( 1  +  a-)  =  wi  ( J  -  -  +  ^  -  ^  +  g-  -  etc), 

the  law  of  which  is  evident.    This  is  the  LogarUhmie  Series,  and 
should  be  fixed  in  memory. 

ScH. — The  Napierian  system  of  logarithms  is  characterized  by  the 
modulus  being  l{m=  1).    Hence  the  Napierian  logarithmic  series  !■ 

/J.8     a^     /j4     /pB 
2+8- 

• 

ISS*  Cor.  1. — Tlie  logarithms  of  the  same  number  in 
different  systems  are  to  each  other  as  the  moduli  of  those 
systems. 

This  is  evident  from  the  general  logarithmic  series.  Thus  the 
logarithm  of  1  +  a;  in  a  system  whose  modulus  is  m,  is  expressed 

a^     a*     ij^     ^ 
log„  (1  +  a')f  =  m(a;  -  g-  +  g-  - 1-  +  g-  -  etc.) ; 

and  the  logarithm  of  the  same  number  in  a  system  whose  modulus  is 
m'  is  expressed 

*  Of  coarse  the  Btadent  will  obeerve  what  forms  the  Bncceedin^;  terms  lo  this 
and  the  other  eimilar  case*  would  have.  Thus  here  we  pboald  have  bF^}i%Gx* 
8-6.7fla;'+  etc. 

t  The  snhscripts  m  and  m'  are  used  to  distintruifh  between  the  pyptemo,  as  k)jf 
(l+x)  \9  not  the  Bame  in  one  pystem  as  in  the  other.  Read  log*  (1  +  x),  "  logarithm 
of  1  -fa;  in  a  system  whose  modolas  is  m,"  etc. 


408  THE    LOGABITTTMTC    SERIKS. 

log„,/(l  +2')  ~m  {x~—  -^  ^  —  -r-  +  -~  etc.). 

i  6  'k  o 

Now,  as  the  number  (1  4- a*)  is,  by  hypothesis,  the  same  in  both  cases. 
X  is  the  same.     Hence,  dividing  one  equation  by  the  other,  we  have 

log,„/(H-a;)""m'' 

166,  Cor.  2,~Having  the  logarithm  of  a  number  in  the 
ISapierian  system^  ive  have  but  to  multiply  it  by  the  moduh(s 
of  any  other  system  to  obtain  the  logarithm  of  the  same  num- 
ber in  the  latter  system. 

Or,  the  logarithm,  of  a  number  in  any  system  divided  by 
the  logarithm  of  the  same  number  in  the  Napierian  system, 
gives  the  modulus  of  the  former  system. 

167 •  Prob. — To  adopt  the  Napierian  logarithmic  series 
to  numerical  computation  so  that  it  can  be  conveniently  used 
for  computing  the  logarithms  of  numbers. 

x'^        iT^        X^       3^ 

Dem.— That  log  (1  +  ^)  =  ^'  ~  o"  +  o"  ~  r  "^  ^  "^  ®*^-'  ^®  ^°*  ^^  ^ 
practicable  form  for  computing  the  logarithms  of  numbers  will  be 
evident  if  we  make  the  attempt.  Thus,  suppose  we  wish  to  compute 
the  logarithm  of  3.     Making  x  =  2,  we  have  log  (1  +  2)  =  log  3  =  2— 

22      2'      2*     2^  ,     ,      , 

—  +  rz —  7"  +  ^ —  6*c.,  a  series  m  which  the  terms  are  growing 

2       3       4       5 

larger  and  larger  (a  diverging  series). 

We  wish  a  series  in  which  the  terms  will  grow  smaller  as  we 
extend  it  (a  converging  series).  Then  the  farther  we  extend  the 
series,  the  more  nearly  shall  we  approximate  the  logarithm  sought. 
To  obtain  such  a  series,  substitute  ~x  for  x  in  the  Napierian  loga- 
rithmic series,  and  we  have 

log  (1-05)  = -ar-- ------- etc. 

Subtracting  this  from  the  former  series,  we  have 
log  {l  +  x)-\o^a-x)  =  log  (~^\  =2(x  +  {a^  +  ^^  +  \x''  +  etc.). 


THE    LOGAlilTHMIC    SERT15S. 


40ft 


1 
+  1' 

■X  Z 

substituting,  and  transposing, 
log(l+z)=rlog24-2^^— 


Now  put  X  = 

-. — -  ,  and    7 — 
22-1-1  1- 


whence   l  +  a-  =  1  +  ^r-^-,  =?— ?.   l-x  = 
2e-l-l      2«  +  l 

Hence,  as  log  ( — -J  =  log  (1  +«)  —  log  z. 


1 


1 


'd(2z  +  lf     5,2z-fl)»      7(22 +  iy 


+  etc. 


)(A) 


This  series  converges  quite  rapidly,  especially  for  large  values  of 
z,  and  is  convenient  for  nse  in  computing  logarithms. 

lOS,  I^rob, — To  compute  the  Napierian  logarithms  of 
the  natural  numbpyn  1,  2,  3,  4,  etc^  ad  libitum. 

Solution.— In  the  fii-st  place  we  remark  that  it  is  only  necessary 
to  compute  the  logarithms  of  prime  numbers,  since  the  logarithm  of 
a  composite  number  is  equal  to  the  sum  of  the  logarithms  of  its 
factors  {121). 

Therefore  beginning  with  1,  we  know  that  log  1=0  (ScH.,  p.  382). 

To  compute  the  logarithm  of  2,  make  2  =  1,  in  series  (A),  and  we 

h«ve  log  (1  +  1)  -  ,„g  1  =  log  2  =  2  (1  +  3I,  +  _^.  +  _1_  +  _>_  ^ 

1  1  1  \ 

il.3>'"*'l3.3'»"^15.8»^"^  ^^^7 

Th«  numerical  operations  are  conveniently  performed  as  follows: 


8 

200000000 

0 

.66666(567 

1 

.66666667* 

9 

.07407 '07 

8 

.02469136 

0 

.0082804-. 

5 

.00164603 

9 

.00001449 

7 

.00013064 

9 

.000 10 1*^1 

9 

.00001129 

9 

.00001129 

11 

.00000103 

9 

.00000125 

18 

.00000009 

.00000014 

15 

.00000001 

.-.  l0| 

?2  = 

:  .69314718  • 

♦  Though  the  declmnl  part  of  n  lojrarithra  Is  penerally  not  exact,  It  1*  not  cus- 
tomary to  annex  the  +  sipn. 


410 


THK    LOGARITHMIC    SKRIES. 


Second.    To  find  log  'S,  make  2  —  2,  whence 

log  3  =  log  2+2  Q  +  gi^,  +  _^  +  Jg,  +  -Ig,  ^  etc.). 


COMPUTATION. 

5 

2.00000000 

25 

.40000000 

1 

.40000000 

25 

.01600000 

3 

.00533333 

25 

.00064000 

5 

.00012800 

25 

.00002560 

7 

.00000366 

.00000102 

9 

.00000011 

.40546510 

log 

2  = 

.69314718 

/.  log  3  =  1.09861228 

Thikd.    To  find  log  4. 

Log  4  =  2  log  2  =  2  X  .69314718  =  1. 

FouETH.     To  find  log  5.    Let  x  =  4,  whence 

log  5  =  log  4  +  2  Q  +  ^g-3  +  Jj;  +  ^,  +  etc.). 


coMPm 

'ATIO 

N. 

^ 

2.00000000 

81 

.22222222 

1 

.22222223 

81 

.00274348 

3 

.00091449 

81 

.00003387 

5 

.00000677 

.00000042 

7 

.00000006 

.22314354 

log  4  =  1.38629430 

•*•  log 

5  = 

1.60943790 

In  like  manner  we  may  proceed  to  compute  the  logarithms  of  the 
prime  numbers  from  the  formula,  and  obtain  those  of  the  composite 
numbers  on  the  principle  that  the  logarithm  of  the  product  equals 
the  sum  of  the  logarithms  of  the  factors. 

Thus,  the  Napierian  logarithm  of  the  base  of  the  common  system, 
10,  =  log  5  +  log  2  =  2.30258508. 


HIGHER    EQUATIONS.  411 

lOO,  Prop, — The  modulus  of  the  common  system  is 
.43429448  +  . 

Dem. — Since  the  logarithm  of  a  number,  in  any  system,  divided  hv 
tlie  Napierian  logarithm  of  the  same  number  is  equal  to  the  modulus 
of  that  system  {100),  we  have 

Com.  log  10  A  ^        , 

=^ r     10  ~  nioaulus  of  common  system. 

But  com.  log  10  =  1,  and  Nap.  log  10  =  2  30258508,  as  found 
above.    Hence, 

Modulus  of  common  system  =  Koi^KOKtvo  =  -43429448 

*.oO*OoOUo 


170.  Prop.— The  Napierian  base  is  2.718281828. 

Dem.— Let  e  represent  the  base  of  the  Napierian  system.     Then 
by  (165) 

com.  log  e  :  Nap.  log  e  : :  .43429448  :  1. 

But  the  logarithm  of  the  base  of  a  system,  taken  in  that  system  is  1, 
since  a'  =  a.  Hence,  Nap.  log  e  =  1,  and  com.  log  e  =  .43429448. 
Now  finding  from  a  table  of  common  logarithms  the  number  corres- 
ponding to  the  logarithm  .43429448,  we  have  e  =  2.718281828. 


IMCHOM  I¥. 


HIGHER     EQUATIONS. 

171.  Since  every  equation  with  one  unknown  quantity, 
and  real  and  rational  coefladents,  can  be  transformed  into 
one  of  the  form 

:t^  +  A3^-^  +  Bx^-^-\-Cx''-^ L  =  0,  (1) 

this  will  be  taken  as  the  typical  numerical  equation  whose 
solution  we  shall  seek  in  this  and  the  succeeding  sections; 


412  HIGHER    EQUATIONS. 

and  we  shall  frequently  represent  it  by  /  (x)  =  0,  read 
"function  x  equals  0."  The  notation  f(x)  signifies  in 
general,  as  has  been  before  explained  (157),  simply  any 
expression  involving  x.  Here  we  use  it  for  this  particular 
form  of  expression.  We  shall  also  use/'  {x)  as  the  symbol 
for  the  first  differential  coefiicient  of  this  function. 


172,  J^roiy. —  Wlien  an  equation  is  reduced  to  the  form 

x"-f  Ax"-'4-Bx"-*4-Cx"-^ L  =  0,  the  roots,  with  their 

signs  changedy  are  factors  of  the  absolute  {known)  terrUy  L. 

[For  demonstration  see  p.  363.] 

173.  Cor. — If  a  is  a  root  of  f  (x)  =  0,  f  (x)  is  divisible 
by  X — a ;  a7id,  conversely,  ifi(x)  is  divisible  by  x--a,  a  is  a 
rootofi{x)=zO. 

Dem. — The  first  statement  is  demonstrated  in  the  proposition,  and 
the  second  is  evident,  since  as  f(x)  is  divisible  by  x—a,  let  the 
quotient  be  (l>{x);  whence  {x—a)(t>{x)  —  0.  Now  x  =  a  will  satisfy 
this  equation,  since  it  renders  x—a  —  Q,  and  does  not  render  (p{x) 
Infinity,  since  by  hypothesis  x  does  not  occur  in  the  denominator.* 


174*  l?rop,—If  the  coefficients  and  absolute  term  in 

x°  +  Ax"-'  +  Bx"-'4-Cx"-* L  =  0,  are  all  integers,  the 

equation  can  have  no  fractional  root. 

Dem. — Suppose  in  this  equation  x  :=  -  \  -  being  a  simple  fraction 

V  t 

in  its  lowest  terms.     Substituting  this  value  of  x,  we  have 

jtN  a>(— 1  iM—i  on — 3 


*  Could  there  be  a  term  of  the  form  —  in  <p  (x),  x  =  a  would  render  It  oc,  and 

X — a 

ax— a)  <p  (x)  would  be  Ox  ao,  which  1?  indeterminate,  since  0x3o  =  0xJ  =  §. 


HIGHER    EQUATIONS.  413 

Multiplying  by  t— »  we  obtain 

J  +  ^»"-'  +  BUT-*  -H  CV««— ^ ....  Z<^»  «:  a 

Now,  by  hypothesie,  all  the  tenns  except  the  first  are  Integral,  and  the 
first  is  a  simple  fraction  in  its  lowest  terms,  as  by  hypothesis  *  and  t 
are  prime  to  each  other.  But  the  sum  of  a  simple  fraction  in  its  low- 
eat  terras  and  a  series  of  integers  cannot  be  0.    Therefore  x  cannot 

equal  -  ,  a  fraction. 
{ 

77'«>»   ScH. — This  proposition  does  not  preclude  the  possibility  of 

turd  roots  in  this  form  of  equation.    These  are  possible. 


116.  ^Frop. — An  equation  f  (x)  =  0  (171)  of  the  nth 
degree,  has  n  roots  {if  it  has  any),  and  no  more. 

Dem. — Let  a  be  a  root  of  /  {x)  =  0,  which  is  of  the  nth  degree. 
Dividing /(x)  by  a;  —  a  {173),  we  have  f  (x)  =  0,  an  equation  of  the 
(71  —  l)th  degree. 

Let  6  be  a  root  of  <p  (x)  —  0,  and  divide  ^  (a;)  by  a;  —  6  {17S).  Call 
the  quotient  0'  {x),  whence  <p'  {x)  =  0,  an  equation  of  the  (n  —  2)th 
degree.  In  this  way  the  degree  of  the  equation  can  be  diminished  by 
division  until,  after  n  —  \  divisions,  there  results  <^*{x)  of  the  first 
degree,  and  the  equation  is  a;  —  ^  =  0.  Therefore, 
f{x)  ={x-'a)  ip{x)  =  (x-~a){x  —  b)  (jt'ix)  =  (x  —  a)  (x  —  b)  (x  —  e) 

f  (x)  =  (x  —  a)  (a;  —  6)  (a;  —  c) {x  —  I)  =  0  ; 

X.  e.,f(x)  is  resolvable  into  n  factors,  of  the  form  x  —  m. 

Now,  as  a;  =  a,  or  a;  =  6,  or  a;  =  any  one  of  the  quantities  a,  b,  e 
....  I,  will  render /(x)  equal  to  0,  each  one  of  these  will  satisfy  the 
equation /(x)  =  0.     Therefore  this  equation  has  ;i  roots. 

Again,  since  it  is  evident  that  we  have  resolved /(x)  into  its  prime 
factors  with  respect  to  x,  there  can  be  no  other  factor  of  the  form 
x  —  m  in/(x),  hence  no  other  root  of /(x)  =  0,  and  this  whether  m  i.s 
equal  to  one  or  more  of  the  roots  a,  &,  c  .  .  .  .  n,  or  not.  Therefore 
f{x)  =  0  has  only  n  roots. 

177.  Cor.  h—TIie  polynomial  x°  -f  Ax°-'  +  Bxn-'-f  Cx"-' 
....  L,  or  f  (x),  =  (x  —  a)  (x  —  b)  (x  —  c)  .  . .  .  (x  —  1),  in 
toJUeh  a,  b,  0  . ,  .  .  \  are  the  roots  of  i  (x)  r=  0. 


414  HIGHER    EQUATIONS. 

1^8.  Cor.  2. — The  equation  f  (x)  =  0  may  have  2,  3,  or 
even  n  equal  roots,  as  titer e  is  no  inconsistency  in  supposing 
a  =;:  b,  a  =  b  =  c,  or  a  =  b  =  c  =  .  .  .  ,  ly  in  the  above 
demonstration. 

179.  Cob.  3. — Imaginary  roots  enter  into  equations 
having  only  real  coefficients,  in  conjugate  pairs  j  that  is,  if 
f  (x)  =  0  has  only  real  coefficients,  if  it  has  one  root  of  the 
form  «  +  ^V—  1,  it  has  another  of  the  form  «  —  ^V^^l ; 
or,  if  it  has  one  of  the  form  ^V^h  it  has  aiiother  of  the 
form  —§  V—  1. 

This  is  evident,  since  only  thus  can  /  {x)  =  {x  —  a){x  —  b)  {x  —  c) 
. .  .  .  (x  —  n);  that  is,  if  one  root,  a  for  example,  is  a  —  /jy'Zri, 
there  must  be  another  of  the  form  a  +  /3/y/ —  1,  in  order  that  the  pro. 
duct  of  these  two  factors  shall  not  involve  an  imaginary.  Thus, 
[x~{a  +  (3^/:^l]  X  [x  —  (a  —  P^^^l]  =  x'^  —  2ax  +  (a^  +  /32),a  real 
quantity.  So  also  (x  —  (3^ — 1)  (a;  4-/3  V — 1)  =  x"^  +  /?^  a  real  quan- 
tity. But  if  the  assumed  imaginary  roots  be  not  in  conjugate  pairs, 
the  product  of  the  factors  {x  —  a)  {x  —  b)  (x  —  e) .  .  . .  {x  —  I)  will  in- 
volve imaginaries. 

180,  Cor.  4. — ffence  an  equation  of  an  odd  degree  must 
have  at  least  one  real  root  j  hut  an  equation  of  an  even  degree 
does  not  necessarily  have  any  real  root. 

Cor.  5. — If  an  equation  has  a  pair  of  imaginary  roots, 
the  known  quantities  entering  into  the  equation  may  he  so 
varied  that  the  two  imaginary  roots  shall  first  give  place  to 
two  equal  roots,  and  then  these  to  two  unequal  roots. 

As  shown  above,  imaginary  roots  arise  from  real  quadratic  factors 
in/(:c).  Let  x^—  2ax  +  &  be  such  a  quadratic  factor,  whence  x^ —  2ax  + 
b  =  0  satisfies /(.t)  =  0,  and  a  ±  \/a^  —  b  are  the  corresponding  roots 
oif{x)  =  0.  Now,  if  6  <  a'^,  these  roots  are  imaginary.  If,  however, 
b  diminishes  or  a  increases  (or  both  change  thus  together),  when 
b  ^  a^  the  two  imaginary  roots  disappear  and  we  have  in  their  placo. 


HIGHER    EQIATIONS.  416 

two  real  roots,  each  a.  If  the  same  change  in  a  and  b  continues,  so 
that  a*  becomes  greater  than  6,  the  two  real,  equal  roots  in  turn  give 
place  to  two  real,  unequal  r<x)ts.  Now  as  a  and  b  are  functions  of  the 
known  quantities  of  the  equation /(j)  =  0,  such  changes  are  evidently 
possible. 

J81,  By  means  of  the  property  exhibited  in  (/77), 
produce  the  equations  whose  roots  are  given  in  the  follow- 
ing examples : 

1.  Roots  1,  -  3,  4. 

2.  Roots  >v/2,  —  V2,  —  1,  3. 

3.  Roots  1,  2,  2,  —  3,  4. 

4.  Roots  -  3,  2  +  V^^h  2  -  V^^ 

5.  Roots  3,  —  2,  —  2,  —  2, 1. 

6.  Roots  I,  i,  -  {. 

7.  Roots  1  ±  V^,  2  ±  \/^=r§". 
a  Roots  li,  2,  >v/3,  -  V3. 

9.  Roots  \/^^,  -  V^^,  \/6,  -  \/5. 

10.  Roots  10,  -  13,  i,  1. 

11.  Roots  3  -  2  VS,  3  +  2  a/3,  2  -  3  sT^,  2  -f  8 

\/=^,  1,  -  1. 


1S2,  !Prop. — If  the  equation  f  (x)  =  0  has  equal  roots, 
the  highest  common  divisor  o/f{x)  and  its  differential  coeffi- 
cient* f '  (x),  being  put  equal  to  0,  cotistitutes  an  equation 
which  has  for  its  roots  these  equal  roots,  and  no  other  roots.\ 


*  The  differential  coefficient  of  a  function  is  sometimes  called  its  first  derived 
polynomial, 

The  student  must  not  suppope  that  the  roots  ot/(x)  =  0,  and  it«  flr»t  differen- 
tial coefficient/' (sr)  =  0,  are  necessarily  alike,  /'(z)  =  a  series  of  l^rms  some  of 
which  may  be  +  and  some  -,  and  which  may  destroy  each  other,  so  as  to  render 
fix)  =  0,  for  other  values  of  x  than  puch  as  render/ (z)  =  0,  and  not  neoeM»rUy 
for  any  which  do  render/  («)  =  0,  except  the  equal  roots  of  the  latter. 


416  HIGHEE  EQUATIONS. 

Dbm. — Let  a  be  one  of  the  m  equal  roots  of  f{x)  =  0,  and  let  the 
other  roots  be  b,  e  .  . .  .  I;  then  f{oc)  =  {x  —  a}" {x  ^b)  {x  —  c) .  . . , 
{x  —  l)  {177)-    Differentiating  {149)  and  dividing  by  dx,  we  have 

/'  {x)  =  m{x-  ar-^  {x~b){x-e) {X'-l)  +  {x-a)>»{x-e) 

{x  —  I)  +  ....  +  {x  —  ay"{x  —  b)  {x  —  c) .  . .  .  -f  etc. 

Now  {x — a)'""*  is  evidently  the  highest  common  divisor  of  f{x)  and 
f  {x),  and  {x  —  a)^~^  =  0  is  an  equation  having  a  for  its  root,  and 
having  no  other. 

In  a  similar  manner,  if  f{x)  =  0  has  two  sets  of  equal  roots,  so 
that 

f{x)  —  {x  —  ay  {x  —  by  {x  —  c)  («  —  d)  .  .  .  .{x  —  I), 

differentiating  and  dividing  by  dx,  we  have 

f{x)  -  m{x  —  a)"-'  {x  —  by{x-'C){x  —  d) (a?  —  ?)  +  (a;  —  «)"• 

r{x  —  6)'^^  {x  -^  c){x  —  d) . . .  .{x  —  l) 

+{x  —  a)m(x  —  by{x  —  d) .  .  .  .{x  —  n)  +  {x  —  a)m  (x  —  by  {x  —  c) 
. .  . .  {x  —  1)  +  . .  .  .  +  {:x  —  a)"»(.'Z!  —  by{x  —  c)  (.c  —  d) . .  .  .  +  etc. 

Now  the  highest  common  divisor  of  f{x)  and  f\x)  is  evidently 
{x  —  rt)*"-^  (x  —  by-K  Putting  this  equal  to  0,  we  have  (.r  —  a)"*-^ 
{x  —  by~'^  =  0,  an  equation  which  is  satisfied  hy  x  =  a  and  x  —  b,  and 
by  no  other  values. 

Thus  we  may  proceed  in  the  case  of  any  number  of  sets  of  equal 
roots. 

183,  Sen. — In  searching  for  the  equal  roots  of  equations  of  high 
degree,  it  may  be  convenient  to  apply  the  process  of  the  proposition 
several  times.  Thus,  suppose  that  /(ic)  =  0  has  m  roots  each  equal 
to  a,  and  r  roots  each  equal  to  b.  Then  the  highest  common  divisor 
off{x)  and /'(a;)  is  of  the  form  (x  —  a)'^-^  {x  —  by-'^ ;  whence  {x—a)"^^ 
(x  —  5/-1  =  0  is  an  equation  having  the  equal  roots  sought.  Tliere- 
fore  we  can  find  the  highest  common  divisor  of  {x — ay^-^x — b)'-^,  and 
its  differential  coefficient  which  will  be  of  the  form  (,/• — ay^—^v—by  "^ 
and  write  {x  —  a)"*"^  {x  —  by-^  —  0,  as  an  equation  containing  the 
roots  sought.  This  process  continued  will  cause  one  of  the  factors 
{x  —  a)  or  {x  —  6)  to  disappear  and  leave  {x  —  (Cy^'  —  0,  when  m  >  r; 
(x — by~^  =  0,  when  r>m;  or  {x  —  a)  {x  —  b)  z=  0,  when  w  =  n 
From  any  one  of  these  forms  we  can  readily  determine  a  root 


HIGHER    EQUATIONS.  417 

1S4:,  Proj), — 1)1  an  equation  f  (x)  =  0,  f(x)  will 
chauyt  sign  when  x  jjasses  through  any  real  rooty  if  there  ti< 
but  am  such  root,  or  if  there  is  an  odd  7iumber  of  such 
roots  ;  biit  if  there  is  an  even  number  of  such  roots,  f  (x) 
will  not  change  sign. 

Let  a,  b,c  .  .  .  .e  be  the  roots  of  /(^x)  =  0,  so  that  /(j)  =  {x—a) 
u—b)  {x—e) ....  {x-e)  =  0  {177)-  Conceive  x  to  start  with  some^ 
value  less  than  the  least  root,  and  continuously  increase  till  it 
becomes  greater  than  the  greatest  root.  As  long  as  x  is  less  than  the 
least  root,  all  the  factors  a'— a,  x—b,  etc.,  are  negative;  but  when  a- 
passes  the  value  of  the  least  root,  the  sign  of  the  factor  containing 
that  root  will  become  +,*and  if  there  is  no  other  equal  root,  this 
factor  will  be  the  only  one  which  will  change  sign.  Hence  the  pro- 
duct of  the  factors  will  change  sign.  But  if  there  is  an  even  number 
of  roots,  each  equal  to  this,  an  even  number  of  factors  will  change 
sign ;  whence  there  will  be  no  change  in  the  sign  of  the  function. 
If,  however,  there  is  an  odd  number  of  equal  roots,  the  passage 
of  x  through  the  value  of  this  root  will  cause  a  change  of  sign 
in  an  odd  number  of  factors,  and  hence  will  change  the  sign  of  the 
function. 

Finally,  as  it  is  eiident  that  the  signs  of  the  factors,  and  hence  of 
the  function,  will  remain  the  same  while  .r  passes  from  one  root  to 
another,  and  in  all  cases  changes  or  does  not  change  as  above  when  x 
passes  through  a  root,  the  proposition  is  established. 

Th«  following  example  will  be  found  very  instructive  :  .z^  +  4ar*— 
14a!='— 17x— 6  =  0.  The  least  root  of  this  equation  is  — 3.  When  x< 
— 3,/(aj)  is  — ;  when  x  =  —S,f{x)  =  0;  when  x  passes  —3,  increas- 
ing,/(j)  changes  from  —  to  +,  and  remains  +  till  x  =  —I,  when  it 
becomes  0,  and  changes  sign  as  x  passes  —1,  noUcithsta tiding  they  are 
equal  roots.  But  there  i.s  an  odd  number  of  such  roots,  viz.,  three. 
But  in  a:*— 14a;' +  64a'— 96  =  0,  two  equal  roots  of  which  '.re  4,  if  we 
substitute  2  we  get  fix)  =  —16,  and  substituting  5,  f{x)  =  —  1,  the 
function  not  changing  sign,  although  a  root  has  been  passed. 


*  Suppose  e  be  the  lea«t  root,  and  that  e'  iit  the  next  state  of  x  greater  than  c 
then  ff—c  is  +. 


418  HIGHER    EQUATIONS. 

185*  Prop, — Changing  the  signs  of  the  terms  of  an 
equation  containing  the  odd  powers  of  tlie  unknown  quan- 
tity changes  the  signs  of  the  roots, 

DEJvr,— If  x=  a  satisfies  the  equation  af—Ax*+Bx^—Cx  +  D  =  0, 
we  have  a^— Ao^  +  Ba^—  Ca  + 1)  =  0.  Now  changing  the  signs  of  the 
terms  containing  the  odd  powers  of  x,  we  have  x^—Ax'^—Bq^+Gx  + 
D  =  0.  This  is  satisfied  by  a;  =  —a,  if  the  former  is  by  a;  =  a.  For, 
substituting  —a  for  x,  we  have  a^—Aa'^+Ba^—Ca  +  D  =  0,  the  same 
as  in  the  first  instance. 

180,  Cor. — Changing  the  signs  of  the  terms  containing 
the  even  poivers  will  answer  equally  welly  since  it  amounts 
to  the  same  thing  ;  and  if  we  are  carefid  to  put  the  equation 
in  the  complete  form,  cha7iging  the  signs  of  the  alternate 
terms  will  accomplish  the  purpose, 

III. — The  negative  roots  of  x^~-7x  +  Q  =  0,  are  the  positive  roots  of 
—3^  +  7x  +  Q  =  0,  or  of  x^—7x—6  =  0  (0  being  considered  an  even 
exponent) ;  or,  writing  the  equation  ^'^±0.^2— 7a; +  6  =  0,  changing  the 
signs  of  alternate  terms,  and  then  dropping  the  term  with  its  coeffi- 
cient 0,  we  obtain  the  same  result. 

Again,  the  negative  roots  of  a;*^  —  7a^  — 5-r*  +  8-i'^— 133a;-  +  508a5— 
240  =  0,  are  the  positive  roots  of  a'«  +  7.2'5-5.?"^-8.r8-182a;»— 508aj— 240 
=  0,or  of  —x'-lx'  +  53^  +  Sx^  +  lS2.T^  +  506x  +  24Q  =  0. 


187,  JProb. — To  evaluate*  f  (x)  for  any  particular 
value  of  Xj  as  X  z=z  a,  more  expeditiously  than  hy  direct  sub- 
stitution. 

Solution.— As /(a-)  is  of  the  form  a;**  +  ^a;**-' +  ^a;*-^  +  Ca"*-^ . .  X, 
let  it  be  required  to  evaluate  'X*  +  Ax^  +  Bx^  +  Cx  +  I)  tor  x  =  a.  Write 
the  detached  coefficients  as  below,  with  a  at  the  right  in  the  form  of 
a  divisor ;  thus 

♦  TWs  means  to  find  the  valoe  of.  Thus,  suppose  we  want  to  find  the  value  of 
a*— 5a;*  +2a;*— 5a5»  +6a?'— «— 12,  for  x  =  5.  We  might  substitute  5  for  a,  of  course, 
and  accomplish  the  end.  But  there  is  a  more  expeditious  way,  as  the  solution  pf 
this  problem  will  show. 


HIGHER    EQUATIONS.  419 

^A  +B  +C  +D  \a__ 

a  a^  +  Aa  a^ -\- Aa^ -{- Ba  a^  +  Aa' +  Ba^  +  Ca 


a+A     a^  +  Aa  +  B     a^  +  Aa- -\- Ba -\- C     a*-^Aa^  +  Ba;^  +  Ca  +  D 

Having  written  the  detached  coetficients,  and  the  quantity  a  for  which 
/(i")  is  to  be  evaluated  as  directed,  multiply  the  first  coetficieut  1  by 
a,  write  the  result  under  the  second,  and  add,  giving  a-hA.  Multiply 
this  sum  by  a,  write  the  product  under  the  third  coefficient  B,  and 
add,  giving  a'  +  ^a  +  5.  In  like  manner  continue  till  all  the  coeffi- 
cients (including  the  absolute  term,  which  is  the  coefficient  of  sfi) 
have  been  used,  and  we  obtain  a^  +  Aa^  +  Ba^  +  Ca  +  D,  which  is  the 
value  off(x)  for  x  =  a. 

Illustration. — To  evaluate  x^  —5a;'  +  Sar*  —  3a^  +  6'  —a;— 12,  for 
r  =  5: 

1     _5  4-2  -3  +6  -1  -13  |_5 

_6  _0  JO  ^  205  1020 

0  2  7  41  204  1008 

Now  1008  is  the  value  of  a-*— 6a^  +  2a:*— 3a:«  +  6a-'-x— 12,  for  a;  =  6; 
and  it  is  easy  to  see  that  much  labor  is  saved  by  this  process. 

We  are  now  prepared  for  the  solution  of  the  following 
important  practical  problem : 

1S8»  Prob, — To  find  the  commensurable  roots  of  numer- 
ical higher  equations. 

The   solution  of    this    problem   we  will    illustrate   by    practical 
examples. 


EXAMPLES. 

1.  Find  the  commensurable  roots  of  a^— 27^^102? -\-^a?-^ 
C8a:+4:8  =  0,  if  it  has  any. 

Solution.— By  {174),  if  this  equation  has  any  commensurable 
roots  they  are  integral ; — it  can  have  no  fractional  roots. 

Again,  by  {172),  the  roots  of  this  equation  with  their  signs 
changed  are  factors  of  48.  Now,  the  integral  factors  of  48  are  1,  2,  8, 
4,  6,  8,  12,  16,  24,  48.    Hence,  if  the  equation  has  commensurable 


420  HIGHER    EQUATIOIS^S. 

roots,  they  are  some  of  these  numbers,  with  either  the  -f  or  —  sign. 
We  will,  therefore,  proceed  to  evaluate  f{x)  (i.  c  in  this  case  jfi  — 
3a^-15a^  +  8a;-^  +  68;r  +  48),  for  *■  =  +1,  a;  =  -1,  x  =  +3,  a;  =  -3,  etc., 
\iy{187),  as  follows- 

1    -2  -15  +8  +68  +48  [  +1 

_J  ^  -16  -  8  60 

-1  -16  -  8  60  108 

Hence  we  see  that  f or  a;  =  + 1,  f{x)  =  108,  and  + 1  is  not  a  root  o^ 
f{x)  =  0.    Trying  x=  —1,  we  have 

1 


-3 

-15 

+  8 

+  68 

+  48 

-1 

3 

13 

-30 

-48 

-3 

-12 

30 

48 

0 

Thus  we  see  that  for  x  =  —  l,/(a;)  =  0,  and  hence  that  —1  is  a  root 
of  our  equation. 

We  might  now  divide /(;i')  by  x  +  1  (i  75)  and  reduce  the  degree 
of  the  equation  by  unity.  But  it  will  be  more  expeditious  to  proceed 
with  our  trial.     Let  us  therefore  evaluate /(a;)  for  a;  =  +3.    Thus  : 


1     -3 
0 

-15           +8           +68 

0           -30           -44 

-15           -33              34 

+  48  1  +3 
+_48 

96 

ac©  for  X  — 
have 

+  3, 

,f{x)  =  96,  and  +3  is  not  a  root 

Trying  « 

1     -3 
-3 
-4 

-15            +8            +68 

8               14           -44 

-  7              33               34 

+  48  1  -2 
-48 
0 

Hence  for  x  =  —2,  f{x)  =  0,  and  —3  is  a  root.     Trying  x  =»  +8^ 
we  have 


1     -2 

-15 

+  8 

+  68 

+  48  1  +8 

8 

3 

^36 

-84 

-48' 

1 

-12 

-28 

-16 

0 

*  Of  course  it  Is  not  necessary  to  retain  the  -i:  sign,  as  we  have  don«  In  the  pre- 
eeding  operations :  it  has  been  done  simply  for  emphasis. 


moHEB    EQUATION.^.  421 

Hence  for  r  =  ^3,  /(t)  —  0,  and  +3  is  a  root.      Trying  a;  =r  —  3, 
wa  have 

1     -a  •-^15  +8  +68  +  48  I  -8 

-3  _15  _^  -24  -132 

-5  0  8  44-84 

Hence  for  x  =  ^S,f{x)  =  —84,  and  —3  is  not  a  root.    Trying  a;  =  4, 
we  have 

1     -2  -15  -f  8  +68  +48  |_4 

_4  _8  -28  -80  -48 

2  -  7  -20  -12  0 

Hence  for  x  =  ^,f{x)  =  0,  and  4  is  a  root. 

We  have  now  found  four  of  the  roots,  viz.,  —1,  —2,  3,  and  4. 
Their  product  with  their  signs  changed  is  24.  Hence,  by  (17^)  48+ 
24  =  2  is  the  other  root  with  its  sign  changed,  i.  €.,  there  are  two 
roots  —2. 

That  our  equation  had  equal  roots  could  have  been  ascertained  by 
the  principle  in  {1&2) ;  but  as  the  process  of  finding  the  H.  C.  D.  is 
tedious,  it  is  generally  best  to  avoid  it  in  practice. 


to  11.  Find  the  roots  of  the  following : 

-      2.  ir*— a:S— 39ic24.24a;  +  180  =  0; 

3.  3^+bx^—dx—^5  =  0; 

4.  ic3+2.r2-23a;— G0  =  0; 

6.  a:*— 3x8_i4a;2-|-48a:— 32  =  0; 

6.  ^— 8arJ+13:r— 6  =  0; 

7.  x*—Uxi-\-lSx—S  =  0;* 

9.  afi—ldx*-\-eW-'17lTi  +  2l63'-10S  =  0; 

10.  a:*— 452^— 40.rH-84  =  0; 

11.  a:5-3a:*— 9a*+21a«— 10a;+24  =  0. 


*  In  order  to  apply  the  process  of  evalaation,  the  coefficients  of  the  missing 
powers  must  be  supplied.    Thus  we  have  1  +  0—11  + 18—8. 


4:'22  HIGHER    EQUATIONS, 

12  to  17.  Apply  the  process  for  finding  equal  roots  (192, 
188)  to  the  following: 

12.  2;3-f82:2  4.20a:-hl6  =  0; 

13.  x^—x^—^x^VZ  =  0; 

14.  2)3—50:2—8.^  +  48  =  0; 

15.  ir4_ii^2_|_i8^_8=3:0; 

16.  a:4  4.i3^4.33^2_^31^^10  =  0; 

17.  3?—ldx^-\-%1x^—inx^+^\(jx—10%  =  {). 

18  to  24.  Having  found  all  but  two  of  the  roots  of  each 
of  the  following  by  (187),  reduce  the  equation  to  a  quad- 
ratic by  {17S),  and  from  this  quadratic  6nd  the  remaining 
roots : 

18.  x^—6x^-\-10x—8  ==  0; 

19.  x^—4:ci^—8x  +  l]2  =  0; 

20.  a^—3x^-\-x-\-2  =  0; 

21.  2.4—6.^3  +  240:— 16  =  0; 

22.  o:*—12o:8 +  50.2-2— 84.T  + 49  =  0  ;* 

23.  o:4_9a:8^i7^.2+.27o:— 60  =  0; 

24.  a:«— 40:4— 16.t3+112;?;2—208o:  + 128  =  0; 


25.  2o:8_3^2_|_ 2.^-3  =  0;t 

26.  30,-8— 22-2— 60: +  4  =  0; 

27.  8o.^-26o;2+llo:  +  10  =  0; 

28.  or*— Jo;  +  f\  =  0;     (Look  out  for  equal  roots.) 

29.  x^—Gs^+9ix^—dx-{-i^  =  0, 


Apply  the  method  for  finding  equal  roots. 


t  We  have  x'—'^x''+x  —  ^  =  0.    Pnt  a;  =  |,  whence  J^  —  ^r^y'  +  vy—^  =0, 


or  y»  —  —  y"  +  A'y  —  —  =  ^-  ^^  ^^^  k  =  2,  we  have  y'—3y''+4y—12  =  0,  which 
can  he  solved  as  before,  for  one  value  of  y,  and  the  equation  then  reduced  to  a  quad- 
ratic and  solved  for  the  other  values.  Finally,  rememhering  that  x  =  ^y,  we  have 
the  valaes  of  x  reqnirocL 


INTERPRETATION    OF    EQUATIONS.  4^5 


J^ECTOON  V. 


DISCUSSION,  OR   INTERPRETATION,  OF  EQUATIONS. 

ISO,  To  DiscusSf  or  Interpret^  an  Eqvafiou 
or  an  Aff/ebraic  Expression ,  is  to  determine  its 
significance  for  the  various  values,  absolute  or  relative,  wiiich 
may  be  attributed  to  the  quantities  entering  into  it,  with 
special  reference  to  noting  any  changes  of  value  which  give 
changes  in  the  general  significance. 

Such  discussions  may  be  divided  into  two  classes:  1st. 
The  discussion  of  equations  or  expressions  with  reference 
to  their  constants ;  and  2d.  The  discussion  of  equations  or 
expressions  with  reference  to  their  variables. 

The  following  principles  are  of  constant  use  in  such  dis- 
cussions : 

100.  I*roj), — A  fraction,  when  compared  with  a  finite 
quantity,  becomes: 

1.  Equal  to  0,  when  its  7iumerator  is  0  a7id  its  denomina- 
tor finite,  and  tvhen  its  numerator  is  finite  and  its  denomi- 
nator 00  . 

2.  Equal  to  oo ,  when  its  numerator  is  finite  and  its  de- 
nominator 0,  and  lohen  its  numerator  is  oo  and  its  dejiomi- 
nat or  finite, 

3.  It  assumes  an  indeterminate  form  when  numerator  and 
denominator  are  both  0,  and  when  they  are  both  oo  .* 

Dem. — These  facts  appear  when  we  consider  that  the  value  of  a 
fraction  depends  upon  the  relative  magnitudes  of  numerator  and  de- 
nominator. 

*  By  ttiis  it  is  meant  that  -  and  —  may  have  a  variety  of  valnes,  not  that  they 
neoeBsarily  do  have. 


424  tKTE!tf»Rl5TATrois'    OP    JCQrATIONS. 

1 .  Let  a  be  any  constant  and  x  a  variable,  then  the  fraction  - 

a 

diminishes  as  x  diminishes,  and  becomes  0  when  x  is  0.    Again,  the 

a 
fraction  -  diminishes  as  x  increases,  and  when  x  becomes  oo,  i.e., 
X  I  » 

a 
greater  than  any  assignable  magnitude,   -  becomes  less  than  any 

X 

assignable  magnitude  or  infinitesimal,  and  is  to  be  regarded  as  0  in 
comparison  with  finite  quantities.  (See  130  and  148,  Dem.,  and 
foot-note.) 

2.  As  X  increases,  the  fraction  -  increases,  and  hence  when  x  be- 

a 

comes  infinite,  the  value  of  the  fraction  is  infinite.    Also,  as  x  dimin- 

a 
ishes,  the  value  of  -  increases;  hence,  when  x  becomes  infinitely 

small,  or  0,  the  value  of  the  fraction  exceeds  any  assignable  limits, 
and  is  therefore  oo . 

X 

3.  Finally,  if  x  and  y  are  variables,  -  diminishes  as  x  diminishes, 

y 

and  increases  as  y  diminishes.    What  then  does  it  become  when  a?=0, 

0 

and  y  =  01  i.  e.,  what  is  the  value  of  x?     Simple  arithmetic  would 

0 
lead  us  to  suppose  that  ^  was  absolutely  indeterminate,  i.  e.,  that  it 

0 
might  have  any  value  whatever  assigned  to  it,   for    a  =  5,     since 

0 
0  =  5x0  =  0;  7^~'^i  since  0  =  7x0  =  0,  etc.     But  a  closer  inspection 

0 
will  enable  us  to  see  that  the  symboi  x  is  not  necessarily  indetermi- 
nate, or  rather  that  the  expression  which  takes  this  form  for  particu 
lar  values  of  its  components,  has  not  necessarily  an  indefinite  number 
of  values  for  these  values  of  its  components.     Thus,  what  the  value 

X 

of  -  will  be  when  x  and  y  each  diminish  to  0  will  evidently  depend 

upon  the  relative  values  of  x  and  y  at  first,  and  which  diminishes  the 

x 
faster.     Suppose,  for  example,  that    y  —  ^x;    then       =  ■=- ,    Now, 

y      ox 

suppose  X  to  diminish  ;  the  denominator  will  diminish  5  times  as  fast 
as  the  numerator,  and  whatever  the  value  of  x,  the  value  of  the  frao 


INTERPRETATION    OF    EQIATIONS.  425 


X         X 

tlon  will  be  ^.     So  if   y  =  Ta;,  -  =  ^,  which  is  \  for  any  value  of  sc 

Hence,  when    a:  =  0,    and    y  =  0,     we  have    -  =  ;;  =  ^-  =  r,    or 

y      sj      ox      o' 

ar      0       a?       1         a?      0 

2/~0~7i~7'  ^'iy~0~  "^^  other  value  depending  upon  the  rel- 

X        <X} 

itive  values  of  x  and  y.    So,  also,  if  x  =  oo ,  and  y  =  co,  -  ~  — ;  but 

X      <x>        x       \ 
it  y  =  Qx,  we  liave  ^  =  —  =  ^^  =  ^ .    And  so  if  y  =^  IOj,  we  have 

;p*      00  aj  1 

=  ~  =  Tq-  =  Tq  .    Thus  we  see  that  the  mere  fact  that  numerator 

.  and  denominator  become  0,  or  become  oo ,  does  not  determine  the  value 
of  the  fraction,  i.  e.,  gives  it  an  indeterminate  form. 

191.  A  Real  dumber  or  Quantity/  is  one  which 
may  be  conceived  as  lying  somewhere  in  the  series  of  num- 
bers or  quantities  between  —  oo  and  -f  oo  inclusive. 

hjj, — Thus,  if  we  conceive  a  series  of  numbers  varying  both  ways 
from  0,  i.  «.,  positively  and  negatively  to  «,  we  have 

-  00  ....  -4,  -3,  -2,  -1,  0,   +1,   +2,  +8,  +4,  .  .  .  .  +  oo. 

Now  a  real  number  is  one  which  may  be  conceived  as  situated 
somewhere  within  these  limits;  it  may  be  +  ,  — ,  integral,  fractional, 
commensurable,  or  incommensurable.  Thus,  +15624  and  —15624 
will  evidently  be  found  in  this  series.  +V-  may  be  conceived  aa 
somewhere  between  +5  and  +6,  though  its  exact  locality  could  not  be 
fixed  by  the  arithmetical  conception  of  discontinuous  number.  So, 
also,  —y^  is  somewhere  between  —Sand  —6.  Again,  +  y'5  is  some- 
where between  +2  and  +3,  though,  as  above,  we  cannot  locate  it 
exactly  by  the  arithmetical  conception. 

The  follo^ving  Geometrical  lUvMratum  is  more  complete  than  the 
arithmetical.  Thus,  let  two  indefinite  lines,  as  CD  and  AB,  intersect 
(^cross)  each  other,  as  at  0.  Now  let  parallel,  e<inidistant  lines  be 
drawn  between  them.  Call  the  one  at  a  -f-1,  that  at  b  will  be  +2,  at 
c  +3,  etc.  So,  also,  the  line  at  a'  being  —1,  that  at  h'  will  l>e  —2,  at 
c'  —3,  etc.  Now  conceive  one  of  tliese  lines  to  start  from  an  infinite 
distance  at  the  left  and  move  toward  the  right.     When  at  an  infinite 


i%6 


INTERPKETATION    OF    EQUATIONS. 


distance  to  the  left  of  0  its  value  would  be  —  oo ,  and  in  passing  to  0 
it  would  pass  tbrough  all  possible  negdtim  values.  In  passing  0  it 
becomes  0  at  0,  changes  sign  to  +  as  it  passes,  and  moving  on  to 


.B 


infinity  to  the  right,  passes  through  all  possible  positive  values.  Hence 
we  see  how  all  real  values  are  embraced  between  —  O)  and  +  oo,  in- 
clusive.* 

192,  Ail  Imaginary  Number  or  Quantity  is 

one  which  cannot  be  conceived  as  lying  anywhere  between 
the  limits  of  —  oo  and  +  oo ,  as  explained  above.  The 
algebraic  form  of  such  a  quantity  is  an  expression  involving 
an  even  root  of  a  negative  quantity. f 


EXAM  P  LES 


1.  "What  are  the  values  of  x  and  y  in  the  exprestdons 

aV  ^  a'h       ,        , 
y  =         -   J- ,  when  0 


V  and  «  and  a  are 


a  —a 


a  —  a 


*  For  example,  the  student  who  is  acquainted  with  the  elements  of  geometry 
knowp  how  to  construct  a  line  which  is  exactly  equal  to  4^5  (Geom.,  Part  1, 110). 
rhis  line  ho  can  locate  between  +2  and  +3,  and  also  between —2  and —3,  since 
4/5  is  both  +  and-. 

t  Transcendental  functions  afford  other  forms  of  imaginary  expressions ;  for 
example,  sin-'  2,  sec  ^  X,  log  (—120),  log  (— w),  etc.  But  our  limits  forbid  the  con- 
sideration of  the  interpretation  of  imaginaries,  except  in  the  most  restricted  sense, 
RB  indicating  incompatibility  with  the  arithmetical  sense  of  the  problem. 


INTERPRETATION    OF     EQUATIONS.  427 

unequal  ?  When  b  =z  b'  and  a  =  a' ?  Wlien  a  =  a'  and 
b  and  b'  are  unequal  ?  What  are  the  signs  of  2;  and  y  when 
i  >  ^  and  a  >  a',  the  essential  signs  of  a,  a',  J,  and  h' 
being  +  ?  When  b  y  b'  and  a  <  a'  ?  If  a'  and  ^  are 
essentially  negative,  and  a  =  a',  and  b  =  b',  what  are  the 
values  of  x  and  u?    If  a'  and  Z*'  are  each  0  ? 

2.  What   general    relation   between    a   and    a'  renders 
^  ~-?;  =  0  ?    What  renders  it  oo  ? 


1  +  a« 

Solution. — To  render  :; ,  =  0,  we  must  have  a'— a  =  0,  and 

1  +  aa' 

i  +  aa'  finite  or  infinite;  or  else  we  must  have  \-\-aa'  =  <x>,  while 

a'— a  is  finite  or  0  {li}0).     Now  a— a  =  0  gives  a'  =a;   whence, 

r  = :; ,,  which  is  0  for  any  value  of  a,  finite  or  infinite. 

1  +  aa'      1  +  a"^  '' 

Hence  the  relation  a'  =  a  fulfills  the  first  requirement.    Let  us  now 

see  if    1+aa' =  CO    will   also   fulfill  this  requirement.    This  gives 

aa'  =  00 ,  since  subtracting  1  from  oo  would  not  mnke  it  other  than  cc . 

CO 

Thus  we  have  a'  =  —  .    Hence,  for  all  finite  values  of  a  (including  0) 

a'  is  00 ,  and  j- :  =  — ;    =  - :  which  can  only  be  0  when  o  =  oo . 

1  +  aa'      oaf        a  "^ 

Therefore  the  particular  values  a'  =  <x>  =  a  =  cc,  render  :j — —  =  0 ; 

but  no  general  values  do,  unless  a  =  a'. 

Again,  in  order  that  :; j  =  oo  ,  we  most  have  1  +  aa'  =  0,  and 

°  1  +  aa' 

a'— a  finite  or  infinite;  or  else  we  must  have  a' — a  =  oo,  and  l  +  arf 

finite  or  0.     Now  1  +aa'  =  0  gives 

,     1^ 

___!_       a'— a  _^  "^a'  _a"'  +  l  _a'^+l  _ 

*^"      a''    l  +  aa'~        a/^"  a'—n' "     0      ~^' 

a' 


•  This  reduct\on  Is  made  by  dropping  a  and  1,  since  the  subtraction  of  a  finite 
from  an  inflnile,  or  the  addition  of  a  finite  to  an  infinite,  does  not  change  the  char- 
acter of  the  infinite,  Thuf>,  in  thic  cape,  to  apsume  that  dropjilng  a  and  1  afl'ccted 
the  relation  between  numerator  and  denominator,  would  be  to  assign  to  a  and  1 
some  vaiuea  with  respect  to  the  infinite  a\  Dut  tl.is  is  contrary  tu  the  dcfinitioD  of 
an  Inftnits. 


428  INTERPRLTATION    OF    EQUATIONS. 

for  any  value  of  a',  finite  or  infinite.     Therefore  tlie  general  relation 

a  =5 ,  between  a  and  a'  renders  :; :  =  00.*    Let  us  now  see  if 

a  1  +  aa' 

the  relation  a' — a  =  00  will  do  the  same.    Now  if  a' — a  =  00 ,  one  or 

the  other  {a!  or  a)  must   be   00 .     Let    a'  =  00 .    We   then   have 

~^  =  — ;  =  -  ,  which  can  only  be  co  when  a  =  0.    Hence  the  par- 
1  +  aa'      aa'      a 

iicular  values  a'  =  cx)  and  a  =  0  render  :; ,  =  00 ,  but  no  general 

1  +  aa'  ^ 

values  meet  the  requirement  unless  a  = ^ . 

3.  What  general    relation    between    a    and    a'    renders 
I  -^  aa' 


a'  -\-  a 


=  0  ?    What  renders  it  00  ? 


4.  In  the  expression  y  =  —  2a:  -f  4  ±  v^-— 4x  —  5, 
how  many  values  has  ^,  in  general,  for  any  particular  value 
of  aj  ?  For  what  value  or  values  of  x  has  y  but  one  value  ? 
For  what  values  of  a;  is  y  real  ?  For  what  imaginary  ?  For 
what  values  of  a?  is  ^  positive  ?    For  what  negative  ? 

Solution. — Writing  the  expression  thus, 

we  see  that  the  value  of  y  is  made  up  of  two  parts,  viz.,  a  rational 
part  —  k%x — 4),  and  a  radical  part  y'a;"^ — 4r — 5.  But  the  radical  part 
may  be  taken  with  either  the  -f  or  the  —  sign.  Hence,  in  general, 
for  any  particular  value  of  x  there  are  two  values  ofy.  2d.  But  if  such 
a  value  is  given  to  x  as  to  render  the  radical  part  0,  for  this  value  of  x, 
y  will  have  but  one  value,  viz.,  the  rational  part.  But  the  condition 
^x^—\x—^=  0  gives  a;  =  5  and  — L  Thus,  for  a;  =  5,  y  =  —6,  but 
(me  value;  and  for  x  —  — 1,  y=  +6,  also  but  one  value.  8d.  To 
ascertain  for  what  values  of  x,  y  is  real,  we  observe  that  y  is  real  when 
3^ — 4a; — 5  is  positive,  and  imaginary  when  x^—Ax — 5  is  negative. 
Now  for  X  positive,  a;'— (4a;+5)  is  +  when  a;'  >4r  +  5 ;  and  for  x  nega- 

*  It  is  to  be  observed  that  the  relation  a  = ,  requires  that  a  and  a'  have  dif- 
ferent eseential  signs  ;  while  the  relation  a^  =  a  requires  that  ihey  have  the  8am« 
essential  sij^nB' 


INTERPRETATION     OF     EQUATIONS.  4JW 

live,  we  have  j^-f4r— 5,  which  is  positive  when  x^  +  4i>5.  The 
former  inefiuality  gives  ar' — 4a;4-4>9,  or  x>o;  aud  the  latter  gives 
«*  +  4r + 4  > 9,  or  x>\.  Uenco  lor  positive  values  of  x  greater  than  5, 
y  is  real,  and  for  negative  values  of  x  numerically  greater  than  1,  y  is 
real.  The  4th  inquiry  is  answered  by  this  :  f,  is  imaginary  for  all 
valuesof  a;  between — 1  and  +3.     5th.  To  ascertain  what  +  values  of 


X  render  y  + ,  and  what  — ,  we  observe  that  —  (2a) — 4)  ±  /y/a* — ix — 5 
can  only  be  +  when  the  -♦-  sign  of  the  radical  part  is  taken  and  when 
y^ — kj— 5>2a* — 4.  This  gives  x  <2  ±  \/—'3,  i.e.,  an  imaginary 
quantity.  Hence  y  is  never  -f-  for  a?  + .  Taking?  the  negative  sign  of 
the  radical,  we  see  that  both  parts  of  the  value  of  y  are  — ,  and  conse- 
quently y  is  real  and  negative  for  all  +  values  of  x  which  render  y 
real,  t.  e.,  for  values  greater  than  5.  Finally,  for  x  —  we  have 
y  =  2x4-4  ±  ^x^  +  ix — 5.  Now  when  we  take  the  +  sign  of  the  radi- 
cal, both  parts  are  +  ;  hence  this  value  of  y  is  always  plus.  \Vlien 
we  take  the  —  sign  of  the  radical,  y  is  negative  if  23"  +  4  <  y'a-*  -f  4x — 5. 
But  this  gives  a;<  — 2  ±  y^^.  Hence  y  is  never  negative  for  any 
negative  value  of  x.  Therefore  both  values  of  y  are  positive  and  real 
for  all  negative  values  of  x  numerically  greater  than  1. 

6  to  22.  Discuss  as  above  the  values  of  y  in  the  follow- 
ing; i.  e.,  1st.  Show  how  many  values  y  has  in  general,  and 
whether  they  are  equal  or  unequal;  2d.  For  what  particular 
value  or  values  of  .r,  y  has  but  one  value;  3d.  For  what 
values  of  .r,  y  is  real,  and  for  what  imaginary ;  4th.  For 
what  values  of  a;,  y  is  4-,  and  for  what  —  ;  5th.  Also 
determine  what  values  of  x  render  y  infinite: 

5.  t/2_j_2a:^— 2x^— 4y— a;4-10  =  0;* 

6.  y2—2a;y-f  2^— %4-2a;  =  0; 

7.  y2_|_2xy-fr^— Gy  +  9  =  0; 

8.  f-\-2xy-^Zx^—^x=iO\ 

9.  y'— 2a:y-|-3j2H-2y— 4r— 3ss0; 

10.  y3^2.Ty  — 3:r2— 4:r  =  0; 

11.  y'^—2xy-\-x^  +  x  =  0; 

♦  In  al)  cases  rolve  the  equation  for  y  in  the  flrst  place.    In  th1»  ezampto 


430  INTERPRETATION    OF     EQUATIONS. 

13.  y^—2xy-{-x^-\-2y-{-l  =  0; 

14  i/2_2a;2_2?/  +  6a;— 3  =  0  ; 

15.  y^—2xy—3x^-2i/-{-'7x^l  =  0; 

16.  y^—2xy—2  =  0; 

17.  y^—2xy  +  2y  +  4x^%  =  0\ 

18.  4?/2+4^  +  2^-3a;4-12  =  0; 

19.  3/— 8:c2  =  12  ; 

20.  12«/2-f4a;2  =  20; 

21.  x^  +  y'^=  16; 

22.  ar^— z^2  _  20. 

193.  Arithmetical  Interpretations  of  Nega- 
tive and  Imaginary  Solutions. 

1.  A  is  20  years  old,  and  B  16.  When  will  A  be  twice  as 
old  as  B  ? 

SuG's.— We  have  20  +  ;c  =  3  (16  +  x) ;  whence  a;  =  — 13.  The  arith- 
metical interpretation  of  this  result  is  that  A  will  never  be  twice  as  old 
as  B,  but  that  he  was  twice'  as  old  12  years  ago,  ^.  e.,  when  he  was  8 
andB4. 

2.  A  is  «  years  old,  and  B  b.  When  will  A  be  ^  times  as 
old  as  B?  ror7i>l  what  are  the  possible  relative  values 
of  a  and  h  consistently  with  the  arithmetical  sense  of  the 
problem  ?  Interpret  for  aynb,  a  =:  nb,  a<Cnb  when  w>lc 
Also  for  7^  =  1,  aynb,  a<Cnb,  and  a  =  nb. 

3.  Two  couriers,  A  and  B,  are  traveling  the  same  road  ir 
the  same  direction,  the  former  at  rate  a,  the  latter  at  rate  b 
They  are  at  two  places  c  miles  apart  at  the  same  time 
Where  and  when  are  they  together  ? 

Solution  and  Discussion.— Let  XY  represent  the  road  which  the 
couriers  are  traveling  in  the  direction  from  X  to  Y,  and  A  and  B 
the  stations  which  they  pass  at  the  same  time,  A  being  at  A  when  B 
is  at  B,,  and  P  or  P'  the  place  at  which  they  are  together,     Call  the 


INTEKPRETATION    OF    EQUATIONS.  431 

distance  from  B  to  the  place  at  which  they  are  together  ±x,  -f-j*  when 
D  is  bejond  B,  and  — j  when  it  is  on  the  hither  f*ide  of  A  and  B,  as  at 


sr s TS ■^— 

D  .  Then  the  distance  frrm  A  to  the  point  at  which  they  are  together 
is  e  +  {±x).  Now  disregarding  tlie  essential  sign  of  x,  and  leaving  it 
to  be  determined  in  the  sequel,  we  have 

Distance  A  travels  from  A  =  e  +  x. 
Distance  B  travels  from  B  —       x  ; 

Time  from  passing  A  and  B  to  the  time  they  are  together 

and  r.     But  these  are  equal.     Hence  we  are  to  discuss  the  equation. 

e  +  x      X  be  .  ac 

=  =-,  or  a;  =  — i^ ,  and  c  +  x  =3 z . 

a        0  a—b  a—b 


The  points  to  be  noticed  in  the  discussion  are,  (1)  when  a>b,  (3) 
when  a<b,  (3)  when  a  =  b,  c  being  greater  than  0  in  each  case  but 
not  oo .     Also  the  like  cases  when  c  =  0. 

When  c  <0  but  not  CO . 

We  have,  for  a>b,  x  positive,  which  shows  that  the  point  at  which 
they  are  together  is  at  the  right  of  B,  i,  e.,  in  the  direction  which  they 

X  /       <5  +  X\ 

are  traveling.    The  time-,  t  (or J,  is  positive,  yrliich  shows  that 

they  are  together  after  passing  A  and  B. 

For  a<&,  X  is  negative,  and  c  +  x,  which  equals  -^^»  *»  also  nega- 
tive. This  shows  that  they  were  together  at  a  point  at  the  left  of  A. 
that  is,  before  they  reached  the  stations  A  and  B.  With  this  the 
expreeeions  for  the  time  also  agree.  Thus  .  beoomefl  —  r ,  and  -— 
is  also  negative,  since  in  this  case  «><>. 


432  INTERPRETATION    OF    EQUATIONS. 

---,  .  be       be  ,  ac        ae 

When    a  =  b,    x  —  —  -— -=c3o,    and    c  +  a;  = r  =  -  =  oo 

a—b      0  a—b      0 

which  indicates  that  they  are  never  together. 
When  c  =  0. 

In  this  case  x  =  — \  =  0,  and  c  +  x  =  — -  =  0,  for  a  and  6  unequaL 

a—b  a—b  ^ 

indicating  that  they  are  together  when  they  are  at  A  and  B.    This  is 
evidently  correct,  since  A  and  B  coincide  in  this  case.     When  a  =  b, 

X  —  —  V  =  ;; ,  and  c  4-  a;  =  TT ,  which  shows  that  they  are  always  to- 
a—b      0  0 

gether,  -  being  a  symbol  of  indetermination  which  in  this  instance 

may  have  any  value  whatever,  as  we  see  from  the  nature  of  the 
problem 

194:*  SCH.— The  student  should  not  understand  that  the  symbol 

-  always  indicates  that  the  quantity  which  takes  this  form  has  an 

indefinite  number  of  values.     It  is  frequently  so,  but  not  necessarily. 

The  indetermination  may  be  only  apparent,  and  what  the  value  of  the 

expression  is  must  be  determined  from  other  considerations.    The 

Cal'mlus  affords  the  most  elegant  general  methods  of  evaluating  such 

expressions.     But  the  simple  processes  of  Algebra  will  often  suflBce. 

l_a^      0  1— ar* 

Thus  for  a?  =  1,  z =  x .    But  :; =  1  +x-¥x^,  which,  for  a;  =  1, 

1— a?       0  1— a; 

1— aj* 

is  3.    Hence  :; =  3,  for  .i;  =  1.     Here  the  apparent  indetermina- 

\—x 

tion  arises  from  the  fact  that  the  particular  assumption  (that  x  =  1) 

causes  the  two  quantities  between  which  we  wish  the  ratio,  \vl.,  the 

numerator  and  denominator,  to  disappear.     Let  the  student  find  that 

, L~^      ,  =  2i  for  x  =  \.    (See  also   190,  8d  part  of  demon- 

1 — aj+a^ — ^ 

stration.) 

4.  Two  couriers  starting  at  'he  same  time  from  the  two 
points  A  and  B,  c  miles  apart,  travel  toward  each  other  at 
the  rates  a  and  h  respectively.  Discuss  the  problem  with 
reference  to  the  place  and  time  of  meeting.  (Consider  when 
ayhf  a<,bj  and  a  =  h.) 


IJSTEKPKhlATlu.N    Of    EQUATIONS.  433 

5.  Two  (-ouriers,  A  jiiiil  li,  are  traveling  tlie  same  road  in 
the  same  direction,  tlie  former  at  rate  a,  and  the  latter  n 
times  as  fast  They  are  at  two  places  c  miles  apart  at  the 
same  time.  Discuss  the  problem  with  reference  to  place 
and  time  of  meeting  as  in  Ex.  3,  adding  the  cunsiderations 
n  >  1,  ;/  <  1,  /i  =  1,  71  =  0. 

6.  Divide  10  into  two  parts  whose  product  shall  bo  40. 

Solution  and  Discussion.— Let  x  and  y  be  the  parts,  then  x  +  y 

-  10,  xy  -  40,  and  J?  =  6  ±  V— 15,  y  =  5  t  y/^lS.  These  re- 
sults we  find  to  he  imaginary.  This  signifies  that  the  prohlem  in  its 
arithmetical  signification  is  impossible :  this  indeed  is  evident  on  the 
face  of  it.  But,  although  impossible  iu  the  arithmetical  sense,  the 
values  thus  found  do  satisfy  the  formal,  or  algebraic  sense.  Thus  the 
sum  of  5  +  Y^—  15  and  5  —  >y/  —  15  is  10,  and  the  product  40. 

7.  The  sum  of  two  numbers  is  required  to  be  a,  and  the 
product  b:  wliat  U  the  maximum  value  of  b  which  will  ren- 
der the  problem  pur^sible  in  the  arithmetical  sense  ?  What 
are  the  parts  forthi:  value  of  b  'f 

8.  Divide  a  into  two  parts,  such  that  the  sum  of  their 
squares  shall  be  a  minimum. 

gUG's. — Let  X  and  a  —  j  be  the  parts,  and  m  the  minimum  sum. 
Then 

.r*  +  (a  —  x\^  =  2j^  —  2cu  +  a*^  m; 

whence  x  —  \a  ±  ^  y^/^  -^'.  From  this  we  see  that  if  2m>a'^  x  is 
imaginary.  Hence  the  least  value  which  we  can  have  is  2m  =  a^  or 
m  =  \a?. 

9.  Divide  a  into  two  parts,  such  that  the  sum  of  the 
square  roots  shall  be  a  maximum. 

10.  Let  d  be  the  difference  between  two  numbers:  re- 
quired that  the  square  of  the  greater  divided  by  the  lest 
shall  be  a  minimum, 


434  IN^TERPREIATION   OF    EQUATIONS. 

11.  Let  a  and  h  be  two  numbers  of  which  a  is  the  greater, 
to  find  a  number  such  that  ii  a  be  added  to  this  number, 
and  b  be  subtracted  from  it,  the  product  of  this  sura  and 
this  difference,  divided  by  the  square  of  the  number,  shall 
be  a  marximum. 

Sug's. — Let  n  be  the  number,  and  m  the  required  maximum  quo 

tient.     Then  by  the  conditions „- =  m,  whence  we 

find 


_  a  -h  's/a^  +  ^ab  -\-W  —  4abm 

Prom  this  we  see  that  the  greatest  value  which  m  can  have  and  ren- 

,  ,  .  (a  +  bf      nru-      •  a  —  b  2ab 

der  n  real  is  m  =  ^  ,  ,     .    This  gives  n  =  —  ^r^ \  =  -j . 

Aab  *^  2{l  —  m)      a  —  b 

12.  To  find  the  point  on  a  line  passing  through  two 
hghts  at  which  the  illumination  will  be  the  same  from  each 
light. 

SoiiTJTiON.— Let  A  and  B  be  the  two  lights,  and  XY  the  line  passing 

mmm >- 


— ^ © @ V 

D'  A  B  D 

through  them.  Let  a  be  the  intensity  of  the  light  A  at  a  unit's  dis- 
tance from  it,  b  the  intensity  of  B  at  a  unit's  distance  from  it,  c  the 
distance  between  the  two  lights,  as  AB,  and  x  the  distance  of  the  point  of 
equal  illumination  from  the  light  A,  as  AD  (or  AD').  Then,  as  we  learn 
from  Physics  that  the  illuminating  effect  of  a  light  varies  inversely  as 
the  square  of  the  distance  from  it,  we  have  for  the  illumination  of  the 

point  D  by  light  A  -^ ,  and  for  the  illumination  of  the  same  point  by 

light  B,  /-^;7-^2  •  But  by  the  conditions  of  the  problem  these  effects 
are  equal ;  hence  we  have  the  equation  to  be  discussed ;  viz., 

a  b 

sfi  ^  (fl  —  »?  * 


IXTERPRKTATION    OF    KQIATIONS.  4^)0 


This  gives        ^^^  =  ^  or  ^-Z^  =  ^  ,/ 6  =  ^Vt 
or                        "L-i^t^/I^oTt^VllVi; 
DF,  finally,        a;  =  c  — ^ ,  and  x  =  c        ^ 


which  are  the  values  of  a;  to  be  discussed. 
Discussion.— I.  Let  c  be  finite  an(f>i). 

t  When  a  >b,x  =  c  — ~ >  ic,  since  -     -^ _ ^  1  fnp 

«  >  &.      This  is  as  it  should  be,  since  for  a>  b  the  point  of  equal 
illumination  will  evidently  be  nearer  to  B  than  to  A.    Again,  the  otlier 

value  of  X  gives  x  =  c  — =r^^ ~  >  c,  since  — rr^^ -iB-i-  and  >  1, 

^a  —  \/b  A^a  —  Ajb 

when  a  >b.     Hence  we  learn  that  there  is  a  point  beyond  B,  as  at  D', 

where  the  illumination  is  the  same  from  each  light. 

If  we  assume  \/a  =  2\/6,  AD  =  f  c,  and  AD'  =  2e. 

2.  It  is  evidently  unnecessary  to  consider  the  caee  when  a<b,  since 
this  would  only  situate  the  points  of  equal  illumination  with  ^efe^ 
ence  to  A  as  the  preceding  discussion  does  with  reference  to  B. 

8.  Whena  =  ii 

^a+\/b 

.  a/o,  \/a       . 

since,  ^-^  =  ■  ^     =  f 

'y/a+'\/6      2'y/o 

This  is  as  it  should  be,  since  it  is  evident  that  in  this  case  the  point 
of  equal  illumination  is  midway  between  the  lightj<.  Again,  for  the 
second  value  of  a*,  we  have 

\/a 

X  r=i  C J^ ~  —  00 . 

Aja—\fh 

This  is  also  evidently  correct ;  for  when  the  lights  are  of  equal  inten- 
sity there  can  be  no  point  beyond  B,  for  example,  at  which  the  ilia- 


436  PERMUTATIONS. 

mination  from  A  will  be  equal  to  that  from  B,  except  when  x  =  <x>, 
for  which  the  illumination  is  0  for  each  light.  [Let  the  student  give 
the  reason.] 

II.  When  c  =  0.      In  this  case  the  original  equation  —.  =  -. 

x*      [C—xy 

jeeomes  -r  =  -= ,  whence  a  —  b.     We  then  have 

x=zc  — ^ =  0 ; 

A^a+'y/b 

\/a  c  \/a         0 

and  a;  =  <j  — -J- =  — ^ — -  =  -. 

Va-  \fb       yj  -  ^Jb      0 

The  former  shows  that  there  is  a  point  of  equal  illumination  where 
the  lights  are  (when  c  =  0  they  are  together),  and  the  latter  shows 
that  any  point  in  the  line  is  equally  illuminated  by  each  light.  Both 
these  conclusions  are  evidently  correct 


PERMUTATIONS. 

./9»5.  Combinations  are  the  different  groups  which 
can  be  made  of  w  things  taken  n  in  a  group,  n  being  less 
than  riK 

III. — Taking  the  5  letters  a,  b,  c,  d,  e,  we  have  the  10  following 
combinations  when  the  letters  are  taken  3  in  a  group,  or,  as  it  is 
usually  expressed,  t  ken  3  and  3  :  abc,  abd,  abe,  acd,  a^e,  ade,  bed,  bee, 
bde,  cde.  Taken  3  and  2,  we  have  the  following  10  combinations  :  ab, 
nc,  ad,  ae,  be,  bd,  be,  cd,  ce,  de.  It  is  to  be  noticed  that  no  tir^o  eombina 
Uons  contain  the  same  letters;  i.  e.,  they  are  different  groups, 

196.  Perm  utations  are  the  different  orders  in  which 
things  can  succeed  each  other. 


PKRMUtATIOXi?!.  437 

III. — Tims  the  two  letters  a,  h,  liave  th<'  two  permutations  oft.  h<t. 
The  three  letters  a,  b,  c  have  the  6  i>er!mit«tious  abc,  acb,  c<ib,  hoc, 
bca,  cba. 

197 •  Arranf/enients  are  permutations  of  com- 
binations. 

III. — Taking  the  10  combinations  of  5  letters  taken  3  and  3,  and 
permuting  each  combination,  we  gH  the  arrangements  of  5  letters 
taken  3  and  3.  Thus  the  combination  obc  gives  the  G  arrangements 
abc,  neb,  cab,  bac,  bra,  cba.  In  liko  manner  each  of  the  10  combina- 
tions of  5  letters  taken  3  and  3  will  give  6  arrangements  ;  whence,  iu 
all,  5  letters  taken  8  and  3  have  60  arrangements. 

IDS,  Prop, —  77ie  number  of  Arrangemenfs  of  m  things 
taken  n  and  n  is 

m  (m— 1)  (m-2)  (m-3) (m-n  +  1). 

Dem. — Let  us  consider  the  number  of  arrangements  which  can  be 
made  of  the  m  letters  a,  b,  c,  d,  etc.,  taken  2  and  2.  Letting  a  8t«nd 
first,  we  can  have  ah,  ar,  ad,  etc.,  to  m—1  arrangements.  Letting  h 
stand  first,  we  can  have  ba,  be,  bd,  etc,  to  w— 1  arrangements  in  each 
case,  or  m  (m—1)  arrangements  in  all. 

Again,  each  of  these  in(in  —  1)2  and  2  arrangements  will  give 
171—2  arrangements  3  and  3,  by  placing  before  it  each  of  the  letters 
not  involved  in  it  Thus  we  have  m(w  —  l)  (w— 2)  arrangements  of 
m  letters  taken  3  and  3. 

Once  more,  each  of  these  m{m—l)im—2)Z  and  3  arrangements 
will  give  ?;?— 3  arrangements  4  and  4,  by  placing  In'fore  it  each  of  the 
letters  not  involved  in  it.  Thus  we  have  m{in  —  l)(m—2)(w—'S) 
arrangements  of  m  letters  taken  4  and  4. 

Finally,  we  ol)Berve  the  law ;  t.  e.,  the  number  of  arrangements 
is  equal  to  the  continued  product  of  /w(m  — 1)  (m~2)  (»w— 3) .  ,  .. 
|,w-(n-l)}  or  m(»n-l)  (m-2)  (»n-3) ....  (m-n  +  n 

199.    Cor,   1. —  TIfe  nvwher  of   Pprmntatio7u<  of  m 

thinqs  is 

1.2.3.4...  .m. 

This  is  e\'ident  since  arrangements  become  pf'rmutations  when  the 
number  in  a  group  is  equal  to  the  whole  number  considered ;  ».  e., 
when  n  =  m. 


438  PlfiUMUtATlO^JS. 

200,  OoR.  2. — If  p  of  the  m  letters  are  alike  {as  each  a), 
q  others  alikey  r  others  alike,  etc.,  the  mimher  of  permuta- 
tions is 

l»2.3-4 m 

1P  X  [g  X  [r  X  etc. 

Thus  consider  the  permutations  of  a,  6,  c,  d,  viz.,  cibcd,  bacd,  acdb, 
ycda,  achd,  head,  aide,  lade,  adcb,  bdca,  etc.  Suppose  &  to  become  a, 
then  since  for  any  particular  position  of  c  and  d,  as  in  abed,  there  are 
as  many  permutations  of  the  four  letters  as  there  can  be  permutations 
of  the  two  letters  «  and  b,  viz.,  1x2;  if  b  becomes  a  there  will  be 
1x2  fewer  permutations  when  these  two  letters  are  alike  than  when 

they  are  different,  i.  e.,  • 

1  •  a 

So,  in  general,  if  p  of  the  letters  are  alike,  there  will  be  1  •  2  •  3  .  .  . 
■p,  or  [£  fewer  permutations  than  if  they  are  all  different,  etc. 

201*  Cor.  3. — TTie  number  of  Combinations  of  m  things 
taken  n  a7id  n  is 

m(m--l)  (m— 2)  (m— 3) (m—n-f  1) 

1.2.3-4 n 

Since  arrangements  are  permutations  of  combinations,  the  number 
of  arrangements  of  m  things  taken  n  and  n  is  equal  to  the  number  of 
combinations  of  m  things  taken  n  and  n  multiplied  by  the  number  of 
permutations  of  n  things.  Hence  the  number  of  combinations  is  equal 
to  the  number  of  arrangements  of  m  things  taken  n  and  n  divided  by 
the  number  of  permutations  of  n  things. 


EXAMPLES. 

1.  How  many  permutations  can  be  made  of  the  letters  in 
the  word  marble?  Of  those  in  home?  Of  those  in 
logarithms? 

2.  How  many  arrangements  can  be  made  of  10  colors 
taken  3  and  3  ?  Of  7  colors  taken  2  and  2  ?  Taken  3 
and  3  ?  4  and  4  ?  5  and  5  ?  6  and  6  ?  7  and  7  ?  How 
many  mixtures  in  each  case,  irrespective  of  proportions  ? 


PEBMLTATI0N8.  439 

3.  How  many  ditfereut  products  can  be  made  from  the 
9  digits  taken  2  and  2?  3  and  3  ?  4:  and  4  ?  5  and  5  ? 
6  and  6  ?     7  and  7  ?    8  and  8  ?     9  and  9  ? 

4.  IIow  many  different  numbers  can  be  represented  by 
the  9  digits  taken  2  and  2  ?    3  and  3  ?     1  and  4  ?  etc. 

5.  In  a  certain  district  3  representatives  are  to  be  elected; 
and  there  are  C  candidat-es.  In  how  many  different  wayi 
may  a  ticket  be  made  up  ? 

6.  There  are  7  chemical  elements  which  will  unite  with 
each  other.  lIow  many  ternary  compounds  can  be  made 
from  them  ?    How  many  binary  ? 

7.  How  many  different  sums  of  money  can  be  paid  with 

1  cent,  1  3-cent  piece,  1  5-cent  piece,  1  dime,  1  15-cent 
piece,  1  25-cent  piece,  and  1  50-cent  piece  ? 

Sdg. — If  taken  1  and  1,  how  many  ?  If  2  and  2,  how  manj  ?  If  3 
and  3,  et«.  ?    How  many  in  all  ? 

8.  In  how  many  ways  can  12  ladies  and  12  gentlemen 
arrange  themselves  in  couples  ? 

9.  If  you  are  to  select  7  articles  out  of  12,  how  many 
different  choices  have  you  ? 

10.  How  many  different  sums  can  be  made  from  1.  2,  3, 
4,  5,  6,  taken  2  and  2  ? 

11.  IIow  many  permutations  Ci;n  be  made  from  the 
letters  in  the  word  possessions?  (See  200,)  How 
many  from  the  letters  in  the  word  consistent 
cie  sf 

12.  How  many  different  signals  can  be  made  with  10 
different-colored  flags,  by  displaying  them   1   at  a   time, 

2  at  a  time,  3  at  a  time,  etc.,  the  relative  position  of  the 
flags  with  reference  to  each  other  not  being  taken  into 
account  ? 


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